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A Boy Throws a Ball Straight Up and Then Catches It Again Ignore Air Resistance

Written By miércoles, 18 de mayo de 2022 Add Comment Edit

Learning Objectives

Past the end of this section, you will be able to:

  • Utilise i-dimensional movement in perpendicular directions to analyze projectile motion.
  • Calculate the range, time of flight, and maximum height of a projectile that is launched and impacts a apartment, horizontal surface.
  • Find the time of flying and impact velocity of a projectile that lands at a dissimilar peak from that of launch.
  • Summate the trajectory of a projectile.

Projectile movement is the motion of an object thrown or projected into the air, subject simply to acceleration equally a upshot of gravity. The applications of projectile movement in physics and engineering are numerous. Some examples include meteors as they enter Earth's atmosphere, fireworks, and the motion of whatever ball in sports. Such objects are called projectiles and their path is chosen a trajectory. The motion of falling objects as discussed in Motility Along a Straight Line is a simple 1-dimensional type of projectile motion in which there is no horizontal motion. In this section, we consider two-dimensional projectile motility, and our treatment neglects the effects of air resistance.

The most important fact to think here is that motions along perpendicular axes are contained and thus can be analyzed separately. We discussed this fact in Displacement and Velocity Vectors, where we saw that vertical and horizontal motions are contained. The central to analyzing two-dimensional projectile motion is to break it into 2 motions: i along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible because acceleration resulting from gravity is vertical; thus, there is no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the 10-axis and the vertical axis the y-axis. It is not required that we use this selection of axes; information technology is simply convenient in the instance of gravitational acceleration. In other cases we may cull a different set of axes. (Figure) illustrates the notation for displacement, where we define [latex] \overset{\to }{s} [/latex] to exist the total displacement, and [latex] \overset{\to }{ten} [/latex] and [latex] \overset{\to }{y} [/latex] are its component vectors forth the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, 10, and y.

An illustration of a soccer player kicking a ball. The soccer player's foot is at the origin of an x y coordinate system. The trajectory of the soccer ball and its location at 6 instants in time are shown. The trajectory is a parabola. The vector s is the displacement from the origin to the final position of the soccer ball. Vector s and its x and y components form a right triangle, with s as the hypotenuse and an angle theta between the x axis and s.

Effigy 4.xi The total displacement s of a soccer ball at a betoken along its path. The vector [latex] \overset{\to }{s} [/latex] has components [latex] \overset{\to }{x} [/latex] and [latex] \overset{\to }{y} [/latex] along the horizontal and vertical axes. Its magnitude is south and it makes an angle θ with the horizontal.

To describe projectile movement completely, we must include velocity and acceleration, too as displacement. We must notice their components forth the x- and y-axes. Let's assume all forces except gravity (such as air resistance and friction, for example) are negligible. Defining the positive direction to be up, the components of acceleration are and then very simple:

[latex] {a}_{y}=\text{−}m=-9.eight\,\text{m}\text{/}{\text{s}}^{2}\enspace(-32\,\text{ft}\text{/}{\text{s}}^{2}). [/latex]

Considering gravity is vertical, [latex] {a}_{10}=0. [/latex] If [latex] {a}_{x}=0, [/latex] this means the initial velocity in the x direction is equal to the terminal velocity in the x management, or [latex] {v}_{x}={5}_{0x}. [/latex] With these conditions on acceleration and velocity, we can write the kinematic (Equation) through (Equation) for motion in a uniform gravitational field, including the residue of the kinematic equations for a constant dispatch from Motion with Constant Acceleration. The kinematic equations for motion in a uniform gravitational field become kinematic equations with [latex] {a}_{y}=\text{−}chiliad,\enspace{a}_{x}=0: [/latex]

Horizontal Motion

[latex] {v}_{0x}={five}_{10},\,x={10}_{0}+{v}_{ten}t [/latex]

Vertical Motion

[latex] y={y}_{0}+\frac{1}{two}({five}_{0y}+{v}_{y})t [/latex]

[latex] {v}_{y}={v}_{0y}-gt [/latex]

[latex] y={y}_{0}+{v}_{0y}t-\frac{1}{2}k{t}^{2} [/latex]

[latex] {v}_{y}^{2}={five}_{0y}^{2}-2g(y-{y}_{0}) [/latex]

Using this set of equations, nosotros tin can analyze projectile motion, keeping in heed some important points.

Trouble-Solving Strategy: Projectile Movement

  1. Resolve the motion into horizontal and vertical components along the x– and y-axes. The magnitudes of the components of displacement [latex] \overset{\to }{s} [/latex] along these axes are ten and y. The magnitudes of the components of velocity [latex] \overset{\to }{5} [/latex] are [latex] {v}_{x}=v\text{cos}\,\theta \,\text{and}\,{v}_{y}=v\text{sin}\,\theta , [/latex] where v is the magnitude of the velocity and θ is its direction relative to the horizontal, every bit shown in (Figure).
  2. Care for the motility as ii independent 1-dimensional motions: one horizontal and the other vertical. Use the kinematic equations for horizontal and vertical motion presented earlier.
  3. Solve for the unknowns in the two separate motions: one horizontal and i vertical. Notation that the only common variable between the motions is time t. The problem-solving procedures hither are the aforementioned every bit those for one-dimensional kinematics and are illustrated in the post-obit solved examples.
  4. Recombine quantities in the horizontal and vertical directions to find the total displacement [latex] \overset{\to }{south} [/latex] and velocity [latex] \overset{\to }{v}. [/latex] Solve for the magnitude and direction of the displacement and velocity using

    [latex] s=\sqrt{{x}^{two}+{y}^{ii}},\enspace\theta ={\text{tan}}^{-1}(y\text{/}x),\enspace{v}=\sqrt{{five}_{x}^{ii}+{five}_{y}^{2}}, [/latex]

    where θ is the management of the displacement [latex] \overset{\to }{s}. [/latex]

Figure a shows the locations and velocities of a projectile on an x y coordinate system at 10 instants in time. When the projectile is at the origin, it has a velocity v sub 0 y which makes an angle theta sub 0 with the horizontal. The velocity is shown as a dark blue arrow, and its x and y components are shown as light blue arrow. The projectile's position follows a downward-opening parabola, moving up to a maximum height, then back to y = 0, and continuing below the x axis The velocity, V, at each time makes an angle theta which changes in time, and has x component V sub x and y component v sub y. The x component of the velocity V sub x is the same at all times. The y component v sub y points up but gets smaller, until the projectile reaches the maximum height, where the velocity is horizontal and has no y component. After the maximum height, the velocity has a y component pointing down and growing larger. As the projectile reaches the same elevation on the way down as it had on the way up, its velocity is below the horizontal by the same angle theta as it was above the horizontal on the way up. In particular, when it comes back to y = 0 on the way down, the angle between the vector v and the horizontal is minus s=theta sub zero and the y component of the velocity is minus v sub 0 y. The last position shown is below the x axis, and the y component of the velocity is larger than it was initially. The graph clearly shows that the horizontal distances travelled in each of the time intervals are equal, while the vertical distances decrease on the way up and increase on the way down. Figure b shows the horizontal component, constant velocity. The horizontal positions and x components of the velocity of the projectile are shown along a horizontal line. The positions are evenly spaced, and the x components of the velocities are all the same, and point to the right. Figure c shows the vertical component, constant acceleration. The vertical positions and y components of the velocity of the projectile are shown along a vertical line. The positions are get closer together on the way up, then further apart on the way down. The y components of the velocities initially point up, decreasing in magnitude until there is no y component to the velocity at the maximum height. After the maximum height, the y components of the velocities point down and increase in magnitude. Figure d shows that putting the horizontal and vertical components of figures b and c together gives the total velocity at a point. The velocity V has an x component of V sub x, has y component of V sub y, and makes an angle of theta with the horizontal. In the example shown, the velocity has a downward y component.

Figure 4.12 (a) Nosotros clarify ii-dimensional projectile motion by breaking it into ii independent 1-dimensional motions along the vertical and horizontal axes. (b) The horizontal motility is simple, because [latex] {a}_{x}=0 [/latex] and [latex] {five}_{x} [/latex] is a constant. (c) The velocity in the vertical direction begins to decrease as the object rises. At its highest point, the vertical velocity is nada. Every bit the object falls toward Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x and y motions are recombined to give the total velocity at any given point on the trajectory.

Example

A Fireworks Projectile Explodes High and Away

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/south at an bending of [latex] 75.0\text{°} [/latex] higher up the horizontal, as illustrated in (Effigy). The fuse is timed to ignite the vanquish merely as it reaches its highest betoken above the ground. (a) Calculate the height at which the beat explodes. (b) How much fourth dimension passes between the launch of the shell and the explosion? (c) What is the horizontal displacement of the trounce when it explodes? (d) What is the total displacement from the point of launch to the highest point?

The trajectory of a fireworks shell from its launch to its highest point is shown as the left half of a downward-opening parabola in a graph of y as a function of x. The maximum height is h = 233 meters and its x displacement at that time is x = 125 meters. The initial velocity vector v sub 0 is up and to the right, tangent to the trajectory curve, and makes an angle of theta sub 0 equal to 75 degrees.

Figure 4.13 The trajectory of a fireworks beat out. The fuse is set to explode the crush at the highest point in its trajectory, which is found to be at a pinnacle of 233 m and 125 m away horizontally.

Strategy

The motion tin exist broken into horizontal and vertical motions in which [latex] {a}_{x}=0 [/latex] and [latex] {a}_{y}=\text{−}g. [/latex] We can then define [latex] {x}_{0} [/latex] and [latex] {y}_{0} [/latex] to be zero and solve for the desired quantities.

Solution

(a) Past "height" nosotros mean the altitude or vertical position y above the starting point. The highest point in any trajectory, chosen the noon, is reached when [latex] {v}_{y}=0. [/latex] Since we know the initial and concluding velocities, equally well as the initial position, we use the post-obit equation to notice y:

[latex] {v}_{y}^{two}={v}_{0y}^{2}-2g(y-{y}_{0}). [/latex]

Because [latex] {y}_{0} [/latex] and [latex] {v}_{y} [/latex] are both zilch, the equation simplifies to

[latex] \text{0}={v}_{0y}^{2}-2gy. [/latex]

Solving for y gives

[latex] y=\frac{{5}_{0y}^{2}}{2g}. [/latex]

At present nosotros must observe [latex] {v}_{0y}, [/latex] the component of the initial velocity in the y direction. It is given past [latex] {five}_{0y}={v}_{0}\text{sin}{\theta }_{0}, [/latex] where [latex] {v}_{0} [/latex] is the initial velocity of lxx.0 m/s and [latex] {\theta }_{0}=75\text{°} [/latex] is the initial bending. Thus,

[latex] {v}_{0y}={v}_{0}\text{sin}\,\theta =(70.0\,\text{m}\text{/}\text{s})\text{sin}\,75\text{°}=67.6\,\text{one thousand}\text{/}\text{s} [/latex]

and y is

[latex] y=\frac{{(67.6\,\text{m}\text{/}\text{s})}^{two}}{2(nine.80\,\text{m}\text{/}{\text{s}}^{ii})}. [/latex]

Thus, we accept

[latex] y=233\,\text{grand}\text{.} [/latex]

Note that because up is positive, the initial vertical velocity is positive, as is the maximum height, but the acceleration resulting from gravity is negative. Annotation also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6-one thousand/s initial vertical component of velocity reaches a maximum top of 233 k (neglecting air resistance). The numbers in this case are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is non completely negligible, and so the initial velocity would have to be somewhat larger than that given to attain the aforementioned pinnacle.

(b) As in many physics problems, there is more than one way to solve for the time the projectile reaches its highest point. In this instance, the easiest method is to use [latex] {v}_{y}={v}_{0y}-gt. [/latex] Considering [latex] {v}_{y}=0 [/latex] at the apex, this equation reduces to simply

[latex] 0={v}_{0y}-gt [/latex]

or

[latex] t=\frac{{v}_{0y}}{one thousand}=\frac{67.6\,\text{m}\text{/}\text{s}}{9.80\,\text{m}\text{/}{\text{due south}}^{2}}=half-dozen.90\text{due south}\text{.} [/latex]

This fourth dimension is also reasonable for large fireworks. If y'all are able to see the launch of fireworks, discover that several seconds laissez passer before the beat explodes. Another way of finding the time is by using[latex] y\,\text{=}\,{y}_{0}+\frac{ane}{2}({v}_{0y}+{v}_{y})t. [/latex] This is left for y'all equally an exercise to complete.

(c) Because air resistance is negligible, [latex] {a}_{x}=0 [/latex] and the horizontal velocity is constant, equally discussed before. The horizontal displacement is the horizontal velocity multiplied by time equally given past [latex] ten={x}_{0}+{v}_{10}t, [/latex] where [latex] {x}_{0} [/latex] is equal to zero. Thus,

[latex] x={v}_{x}t, [/latex]

where [latex] {5}_{x} [/latex] is the x-component of the velocity, which is given by

[latex] {v}_{x}={v}_{0}\text{cos}\,\theta =(70.0\,\text{m}\text{/}\text{south})\text{cos}75\text{°}=18.i\,\text{m}\text{/}\text{s}. [/latex]

Time t for both motions is the aforementioned, and then 10 is

[latex] ten=(xviii.1\,\text{g}\text{/}\text{s})half-dozen.ninety\,\text{s}=125\,\text{chiliad}\text{.} [/latex]

Horizontal movement is a constant velocity in the absence of air resistance. The horizontal deportation found here could be useful in keeping the fireworks fragments from falling on spectators. When the shell explodes, air resistance has a major effect, and many fragments state straight below.

(d) The horizontal and vertical components of the displacement were but calculated, and then all that is needed here is to observe the magnitude and direction of the displacement at the highest signal:

[latex] \overset{\to }{s}=125\hat{i}+233\hat{j} [/latex]

[latex] |\overset{\to }{southward}|=\sqrt{{125}^{two}+{233}^{ii}}=264\,\text{m} [/latex]

[latex] \theta ={\text{tan}}^{-1}(\frac{233}{125})=61.eight\text{°}. [/latex]

Note that the angle for the displacement vector is less than the initial angle of launch. To come across why this is, review (Effigy), which shows the curvature of the trajectory toward the ground level.

When solving (Figure)(a), the expression we found for y is valid for whatsoever projectile motion when air resistance is negligible. Call the maximum elevation y = h. Then,

[latex] h=\frac{{5}_{0y}^{2}}{2g}. [/latex]

This equation defines the maximum height of a projectile above its launch position and it depends only on the vertical component of the initial velocity.

Check Your Understanding

A rock is thrown horizontally off a cliff [latex] 100.0\,\text{m} [/latex] high with a velocity of 15.0 m/southward. (a) Define the origin of the coordinate arrangement. (b) Which equation describes the horizontal motion? (c) Which equations describe the vertical motion? (d) What is the rock'due south velocity at the point of impact?

(a) Choose the meridian of the cliff where the stone is thrown from the origin of the coordinate organization. Although information technology is arbitrary, nosotros typically choose time t = 0 to stand for to the origin. (b) The equation that describes the horizontal motility is [latex] x={ten}_{0}+{v}_{x}t. [/latex] With [latex] {ten}_{0}=0, [/latex] this equation becomes [latex] x={v}_{ten}t. [/latex] (c) (Effigy) through (Figure) and (Figure) describe the vertical motion, only since [latex] {y}_{0}=0\,\text{and}\,{v}_{0y}=0, [/latex] these equations simplify greatly to become [latex] y=\frac{i}{two}({v}_{0y}+{five}_{y})t=\frac{1}{2}{v}_{y}t,\enspace [/latex][latex] {v}_{y}=\text{−}gt,\enspace [/latex][latex] y=-\frac{1}{2}chiliad{t}^{2},\enspace [/latex] and [latex] {v}_{y}^{2}=-2gy. [/latex] (d) Nosotros use the kinematic equations to find the x and y components of the velocity at the indicate of impact. Using [latex] {v}_{y}^{ii}=-2gy [/latex] and noting the bespeak of impact is −100.0 grand, nosotros find the y component of the velocity at bear upon is [latex] {v}_{y}=44.three\,\text{m}\text{/}\text{s}. [/latex] We are given the x component, [latex] {v}_{x}=15.0\,\text{m}\text{/}\text{s}, [/latex] and so we can calculate the full velocity at impact: v = 46.8 m/s and [latex] \theta =71.3\text{°} [/latex] below the horizontal.

Example

Calculating Projectile Motion: Lawn tennis Actor

A tennis player wins a lucifer at Arthur Ashe stadium and hits a brawl into the stands at thirty yard/s and at an angle [latex] 45\text{°} [/latex] above the horizontal ((Effigy)). On its way downwardly, the ball is caught by a spectator 10 thousand above the point where the brawl was hit. (a) Calculate the time it takes the tennis ball to reach the spectator. (b) What are the magnitude and management of the ball'southward velocity at bear on?

An illustration of a tennis ball launched into the stands. The player is to the left of the stands and hits the ball up and to the right at an angle of theta equal to 45 degrees and velocity of v sub 0 equal to 30 meters per second. The ball reaches a spectator who is seated 10 meters above the initial height of the ball.

Figure 4.14 The trajectory of a tennis ball hit into the stands.

Strategy

Once again, resolving this two-dimensional motion into two contained one-dimensional motions allows us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. Thus, nosotros solve for t beginning. While the ball is ascent and falling vertically, the horizontal movement continues at a constant velocity. This example asks for the final velocity. Thus, we recombine the vertical and horizontal results to obtain [latex] \overset{\to }{v} [/latex] at final time t, determined in the first function of the case.

Solution

(a) While the ball is in the air, information technology rises and then falls to a terminal position 10.0 m college than its starting distance. We tin can find the time for this past using (Figure):

[latex] y\,\text{=}\,{y}_{0}\,\text{+}\,{v}_{0y}t-\frac{1}{ii}one thousand{t}^{2}. [/latex]

If we take the initial position [latex] {y}_{0} [/latex] to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity:

[latex] {v}_{0y}={v}_{0}\text{sin}\,{\theta }_{0}=(30.0\,\text{m}\text{/}\text{due south})\text{sin}\,45\text{°}=21.2\,\text{m}\text{/}\text{s}. [/latex]

Substituting into (Effigy) for y gives united states

[latex] 10.0\,\text{m}=(21.2\,\text{yard/s})t-(four.90\,{\text{grand/s}}^{\text{two}}){t}^{2}. [/latex]

Rearranging terms gives a quadratic equation in t:

[latex] (iv.ninety\,{\text{k/s}}^{\text{2}}){t}^{ii}-(21.ii\,\text{m/southward})t+10.0\,\text{1000}=0. [/latex]

Use of the quadratic formula yields t = iii.79 south and t = 0.54 s. Since the ball is at a pinnacle of 10 chiliad at two times during its trajectory—once on the way up and once on the way down—nosotros take the longer solution for the time it takes the ball to reach the spectator:

[latex] t=3.79\,\text{south}\text{.} [/latex]

The fourth dimension for projectile motion is determined completely by the vertical motion. Thus, whatever projectile that has an initial vertical velocity of 21.2 g/due south and lands 10.0 m below its starting distance spends 3.79 s in the air.

(b) We can find the final horizontal and vertical velocities [latex] {five}_{x} [/latex] and [latex] {v}_{y} [/latex] with the use of the outcome from (a). Then, we can combine them to find the magnitude of the total velocity vector [latex] \overset{\to }{five} [/latex] and the angle [latex] \theta [/latex] information technology makes with the horizontal. Since [latex] {5}_{x} [/latex] is constant, we tin solve for it at whatever horizontal location. We choose the starting signal because we know both the initial velocity and the initial angle. Therefore,

[latex] {5}_{x}={5}_{0}\text{cos}{\theta }_{0}=(30\,\text{m}\text{/}\text{due south})\text{cos}\,45\text{°}=21.two\,\text{chiliad}\text{/}\text{s}. [/latex]

The final vertical velocity is given by (Figure):

[latex] {5}_{y}={five}_{0y}-gt. [/latex]

Since [latex] {5}_{0y} [/latex] was found in part (a) to exist 21.2 m/south, nosotros have

[latex] {v}_{y}=21.ii\,\text{m}\text{/}\text{due south}-9.8\,\text{g}\text{/}{\text{s}}^{2}(3.79\,\text{s})=-15.9\,\text{m}\text{/}\text{s}. [/latex]

The magnitude of the final velocity [latex] \overset{\to }{v} [/latex] is

[latex] five=\sqrt{{v}_{x}^{two}+{v}_{y}^{2}}=\sqrt{{(21.2\,\text{m}\text{/}\text{due south})}^{two}+{(\text{−}\,\text{15}.9\,\text{m}\text{/}\text{southward})}^{2}}=26.5\,\text{m}\text{/}\text{s}. [/latex]

The management [latex] {\theta }_{v} [/latex] is plant using the inverse tangent:

[latex] {\theta }_{v}={\text{tan}}^{-1}(\frac{{five}_{y}}{{v}_{x}})={\text{tan}}^{-1}(\frac{21.2}{-15.9})=-53.1\text{°}. [/latex]

Significance

(a) As mentioned earlier, the fourth dimension for projectile move is adamant completely past the vertical motion. Thus, whatsoever projectile that has an initial vertical velocity of 21.two one thousand/southward and lands 10.0 1000 beneath its starting distance spends 3.79 s in the air. (b) The negative angle means the velocity is [latex] 53.i\text{°} [/latex] beneath the horizontal at the point of impact. This result is consequent with the fact that the ball is impacting at a bespeak on the other side of the apex of the trajectory and therefore has a negative y component of the velocity. The magnitude of the velocity is less than the magnitude of the initial velocity we await since it is impacting 10.0 m above the launch elevation.

Fourth dimension of Flight, Trajectory, and Range

Of interest are the fourth dimension of flying, trajectory, and range for a projectile launched on a apartment horizontal surface and impacting on the aforementioned surface. In this case, kinematic equations give useful expressions for these quantities, which are derived in the post-obit sections.

Fourth dimension of flight

We can solve for the time of flying of a projectile that is both launched and impacts on a flat horizontal surface past performing some manipulations of the kinematic equations. Nosotros note the position and deportation in y must be zero at launch and at impact on an even surface. Thus, we set up the deportation in y equal to zip and observe

[latex] y-{y}_{0}={five}_{0y}t-\frac{1}{2}g{t}^{2}=({v}_{0}\text{sin}{\theta }_{0})t-\frac{one}{2}yard{t}^{2}=0. [/latex]

Factoring, we have

[latex] t({v}_{0}\text{sin}{\theta }_{0}-\frac{gt}{2})=0. [/latex]

Solving for t gives usa

[latex] {T}_{\text{tof}}=\frac{ii({v}_{0}\text{sin}{\theta }_{0})}{g}. [/latex]

This is the fourth dimension of flight for a projectile both launched and impacting on a flat horizontal surface. (Figure) does not apply when the projectile lands at a different elevation than information technology was launched, every bit we saw in (Figure) of the tennis player hitting the ball into the stands. The other solution, t = 0, corresponds to the time at launch. The time of flight is linearly proportional to the initial velocity in the y direction and inversely proportional to 1000. Thus, on the Moon, where gravity is one-6th that of World, a projectile launched with the aforementioned velocity as on Earth would exist airborne six times as long.

Trajectory

The trajectory of a projectile can exist found by eliminating the time variable t from the kinematic equations for arbitrary t and solving for y(x). Nosotros take [latex] {x}_{0}={y}_{0}=0 [/latex] so the projectile is launched from the origin. The kinematic equation for ten gives

[latex] ten={5}_{0x}t⇒t=\frac{ten}{{v}_{0x}}=\frac{x}{{v}_{0}\text{cos}{\theta }_{0}}. [/latex]

Substituting the expression for t into the equation for the position [latex] y=({v}_{0}\text{sin}{\theta }_{0})t-\frac{i}{2}yard{t}^{2} [/latex] gives

[latex] y=({v}_{0}\text{sin}\,{\theta }_{0})(\frac{ten}{{v}_{0}\text{cos}\,{\theta }_{0}})-\frac{1}{2}g{(\frac{x}{{5}_{0}\text{cos}\,{\theta }_{0}})}^{ii}. [/latex]

Rearranging terms, nosotros have

[latex] y=(\text{tan}\,{\theta }_{0})10-[\frac{1000}{two{({v}_{0}\text{cos}\,{\theta }_{0})}^{two}}]{x}^{2}. [/latex]

This trajectory equation is of the grade [latex] y=ax+b{x}^{2}, [/latex] which is an equation of a parabola with coefficients

[latex] a=\text{tan}\,{\theta }_{0},\enspaceb=-\frac{g}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}. [/latex]

Range

From the trajectory equation we can too find the range, or the horizontal distance traveled by the projectile. Factoring (Effigy), we have

[latex] y=ten[\text{tan}\,{\theta }_{0}-\frac{g}{ii{({5}_{0}\text{cos}\,{\theta }_{0})}^{2}}10]. [/latex]

The position y is zero for both the launch point and the bear upon point, since nosotros are again because only a flat horizontal surface. Setting y = 0 in this equation gives solutions ten = 0, respective to the launch point, and

[latex] x=\frac{2{v}_{0}^{two}\text{sin}\,{\theta }_{0}\text{cos}\,{\theta }_{0}}{g}, [/latex]

respective to the impact point. Using the trigonometric identity [latex] 2\text{sin}\,\theta \text{cos}\,\theta =\text{sin}two\theta [/latex] and setting x = R for range, we detect

[latex] R=\frac{{v}_{0}^{2}\text{sin}two{\theta }_{0}}{one thousand}. [/latex]

Note specially that (Figure) is valid only for launch and affect on a horizontal surface. Nosotros see the range is directly proportional to the square of the initial speed [latex] {v}_{0} [/latex] and [latex] \text{sin}2{\theta }_{0} [/latex], and it is inversely proportional to the acceleration of gravity. Thus, on the Moon, the range would be six times greater than on Earth for the same initial velocity. Furthermore, nosotros come across from the factor [latex] \text{sin}ii{\theta }_{0} [/latex] that the range is maximum at [latex] 45\text{°}. [/latex] These results are shown in (Figure). In (a) we come across that the greater the initial velocity, the greater the range. In (b), we see that the range is maximum at [latex] 45\text{°}. [/latex] This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is somewhat smaller. It is interesting that the aforementioned range is institute for two initial launch angles that sum to [latex] 90\text{°}. [/latex] The projectile launched with the smaller bending has a lower apex than the higher angle, merely they both have the same range.

Figure a shows the trajectories of projectiles launched at the same initial 45 degree angle above the horizontal and different initial velocities. The trajectories are shown from launch to landing back at the initial elevation. In orange is the trajectory for 30 meters per second, giving a range R (distance from launch to landing) of 91.8 m. In purple is the trajectory for 40 meters per second, giving a range R of 163 m. In blue is the trajectory for 50 meters per second, giving a range R of 255 m. The maximum height of the projectile increases with initial speed. Figure b shows the trajectories of projectiles launched at the same initial speed of 50 meters per second and different launch angles. The trajectories are shown from launch to landing back at the initial elevation. In orange is the trajectory for an angle of 15 degrees above the horizontal, giving a range R of 128 m. In purple is the trajectory for an angle of 45 degrees above the horizontal, giving a range R of 255 m. In blue is the trajectory for an angle of 75 degrees above the horizontal, giving a range R of 128 m, the same as for the 15 degree trajectory. The maximum height increases with launch angle.

Figure 4.fifteen Trajectories of projectiles on level ground. (a) The greater the initial speed [latex] {five}_{0}, [/latex] the greater the range for a given initial bending. (b) The result of initial angle [latex] {\theta }_{0} [/latex] on the range of a projectile with a given initial speed. Annotation that the range is the same for initial angles of [latex] 15\text{°} [/latex] and [latex] 75\text{°}, [/latex] although the maximum heights of those paths are unlike.

Example

Comparing Golf Shots

A golfer finds himself in 2 different situations on unlike holes. On the second hole he is 120 m from the dark-green and wants to hit the brawl 90 grand and permit it run onto the green. He angles the shot low to the ground at [latex] 30\text{°} [/latex] to the horizontal to let the ball gyre subsequently impact. On the fourth hole he is ninety m from the green and wants to let the brawl driblet with a minimum amount of rolling after bear on. Here, he angles the shot at [latex] 70\text{°} [/latex] to the horizontal to minimize rolling afterwards touch. Both shots are hit and impacted on a level surface.

(a) What is the initial speed of the brawl at the second hole?

(b) What is the initial speed of the brawl at the fourth hole?

(c) Write the trajectory equation for both cases.

(d) Graph the trajectories.

Strategy

Nosotros run into that the range equation has the initial speed and angle, so we tin solve for the initial speed for both (a) and (b). When we accept the initial speed, nosotros can use this value to write the trajectory equation.

Solution

(a) [latex] R=\frac{{five}_{0}^{2}\text{sin}\,2{\theta }_{0}}{thousand}⇒{five}_{0}=\sqrt{\frac{Rg}{\text{sin}\,2{\theta }_{0}}}=\sqrt{\frac{90.0\,\text{m}(ix.8\,\text{g}\text{/}{\text{s}}^{two})}{\text{sin}(ii(70\text{°}))}}=37.0\,\text{thousand}\text{/}\text{s} [/latex]

(b) [latex] R=\frac{{five}_{0}^{2}\text{sin}\,two{\theta }_{0}}{g}⇒{5}_{0}=\sqrt{\frac{Rg}{\text{sin}\,2{\theta }_{0}}}=\sqrt{\frac{90.0\,\text{m}(9.8\,\text{m}\text{/}{\text{due south}}^{2})}{\text{sin}(ii(xxx\text{°}))}}=31.nine\,\text{grand}\text{/}\text{s} [/latex]

(c)

[latex] \begin{array}{cc} y=ten[\text{tan}\,{\theta }_{0}-\frac{g}{two{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}x]\hfill \\ \text{Second hole:}\,y=x[\text{tan}\,70\text{°}-\frac{9.8\,\text{m}\text{/}{\text{south}}^{2}}{ii{[(37.0\,\text{chiliad}\text{/}\text{s)(}\text{cos}\,70\text{°})]}^{two}}ten]=two.75x-0.0306{x}^{2}\hfill \\ \text{Fourth hole:}\,y=x[\text{tan}\,30\text{°}-\frac{9.eight\,\text{m}\text{/}{\text{s}}^{2}}{2{[(31.9\,\text{m}\text{/}\text{southward)(}\text{cos}30\text{°})]}^{2}}ten]=0.58x-0.0064{x}^{2}\hfill \cease{assortment} [/latex]

(d) Using a graphing utility, nosotros tin can compare the two trajectories, which are shown in (Figure).

Two parabolic functions are shown. The range for both trajectories is 90 meters. One shot travels much higher than the other. The higher shot has an initial velocity of 37 meters per second and an angle of 70 degrees. The lower shot has an initial velocity of 31.9 meters per second and an angle of 30 degrees.

Figure 4.16 Two trajectories of a golf ball with a range of 90 m. The impact points of both are at the aforementioned level every bit the launch point.

Significance

The initial speed for the shot at [latex] 70\text{°} [/latex] is greater than the initial speed of the shot at [latex] 30\text{°}. [/latex] Note from (Figure) that two projectiles launched at the same speed just at different angles take the same range if the launch angles add together to [latex] ninety\text{°}. [/latex] The launch angles in this instance add to requite a number greater than [latex] ninety\text{°}. [/latex] Thus, the shot at [latex] seventy\text{°} [/latex] has to have a greater launch speed to reach ninety g, otherwise it would land at a shorter altitude.

Cheque Your Understanding

If the two golf game shots in (Figure) were launched at the same speed, which shot would have the greatest range?

The golf shot at [latex] 30\text{°}. [/latex]

When we speak of the range of a projectile on level footing, nosotros presume R is very small compared with the circumference of Globe. If, nonetheless, the range is large, Earth curves away below the projectile and the dispatch resulting from gravity changes management along the path. The range is larger than predicted past the range equation given earlier because the projectile has farther to fall than it would on level ground, as shown in (Figure), which is based on a drawing in Newton'due south Principia . If the initial speed is great plenty, the projectile goes into orbit. Earth's surface drops 5 m every 8000 m. In one southward an object falls 5 g without air resistance. Thus, if an object is given a horizontal velocity of [latex] 8000\,\text{thousand}\text{/}\text{south} [/latex] (or [latex] eighteen,000\text{mi}\text{/}\text{hr}) [/latex] near Earth's surface, it volition become into orbit around the planet because the surface continuously falls away from the object. This is roughly the speed of the Infinite Shuttle in a low Earth orbit when it was operational, or any satellite in a low World orbit. These and other aspects of orbital movement, such as Globe's rotation, are covered in greater depth in Gravitation.

The figure shows a drawing of the earth with a tall tower at the north pole and a horizontal arrow labeled v 0 pointing to the right. 5 trajectories that start at the top of the tower are shown. The first reaches the earth near the tower. The second reaches the earth farther from the tower, and the third even farther. The fourth trajectory hits the earth at the equator, and is tangent to the surface at the equator. The fifth trajectory is a circle concentric with the earth.

Figure 4.17 Projectile to satellite. In each instance shown here, a projectile is launched from a very high tower to avoid air resistance. With increasing initial speed, the range increases and becomes longer than it would be on level footing because World curves away beneath its path. With a speed of 8000 m/s, orbit is achieved.

Summary

  • Projectile motion is the motion of an object subject simply to the dispatch of gravity, where the acceleration is constant, as near the surface of Earth.
  • To solve projectile motion issues, we analyze the motion of the projectile in the horizontal and vertical directions using the one-dimensional kinematic equations for x and y.
  • The time of flight of a projectile launched with initial vertical velocity [latex] {v}_{0y} [/latex] on an even surface is given by

    [latex] {T}_{tof}=\frac{2({v}_{0}\text{sin}\,\theta )}{one thousand}. [/latex]

    This equation is valid just when the projectile lands at the same pinnacle from which it was launched.

  • The maximum horizontal distance traveled by a projectile is chosen the range. Over again, the equation for range is valid but when the projectile lands at the same superlative from which it was launched.

Conceptual Questions

Respond the following questions for projectile motility on level ground assuming negligible air resistance, with the initial angle being neither [latex] 0\text{°} [/latex] nor [latex] 90\text{°}: [/latex] (a) Is the velocity ever zero? (b) When is the velocity a minimum? A maximum? (c) Can the velocity ever exist the aforementioned as the initial velocity at a fourth dimension other than at t = 0? (d) Can the speed always exist the same as the initial speed at a time other than at t = 0?

a. no; b. minimum at noon of trajectory and maximum at launch and impact; c. no, velocity is a vector; d. yeah, where it lands

Answer the following questions for projectile motion on level basis assuming negligible air resistance, with the initial angle being neither [latex] 0\text{°} [/latex] nor [latex] ninety\text{°}: [/latex] (a) Is the acceleration always zero? (b) Is the acceleration ever in the same management every bit a component of velocity? (c) Is the dispatch ever opposite in management to a component of velocity?

A dime is placed at the edge of a table so information technology hangs over slightly. A quarter is slid horizontally on the table surface perpendicular to the edge and hits the dime caput on. Which coin hits the ground first?

They both striking the ground at the same fourth dimension.

Bug

A bullet is shot horizontally from shoulder peak (i.v m) with and initial speed 200 grand/southward. (a) How much time elapses before the bullet hits the ground? (b) How far does the bullet travel horizontally?

a. [latex] t=0.55\,\text{s} [/latex], b. [latex] x=110\,\text{m} [/latex]

A marble rolls off a tabletop one.0 thou high and hits the floor at a point 3.0 m away from the table's edge in the horizontal management. (a) How long is the marble in the air? (b) What is the speed of the marble when it leaves the table'due south edge? (c) What is its speed when information technology hits the floor?

A sprint is thrown horizontally at a speed of 10 1000/southward at the bull's-heart of a dartboard 2.4 m away, equally in the following effigy. (a) How far below the intended target does the dart hit? (b) What does your answer tell y'all nearly how skillful dart players throw their darts?

An aeroplane flight horizontally with a speed of 500 km/h at a superlative of 800 grand drops a crate of supplies (see the following effigy). If the parachute fails to open, how far in front of the release point does the crate hit the ground?

An airplane releases a package. The airplane has a horizontal velocity of 500 kilometers per hour. The package's trajectory is the right half of a downward-opening parabola, initially horizontal at the airplane and curving down until it hits the ground.

Suppose the airplane in the preceding trouble fires a projectile horizontally in its direction of motion at a speed of 300 m/s relative to the plane. (a) How far in forepart of the release point does the projectile hit the ground? (b) What is its speed when it hits the ground?

a., [latex] t=12.eight\,\text{s,}\enspacex=5619\,\text{m} [/latex] b. [latex] {v}_{y}=125.0\,\text{m}\text{/}\text{south,}\enspace{5}_{x}=439.0\,\text{yard}\text{/}\text{s,}\enspace|\overset{\to }{v}|=456.0\,\text{one thousand}\text{/}\text{south} [/latex]

A fastball bullpen can throw a baseball at a speed of 40 m/s (90 mi/h). (a) Assuming the pitcher can release the ball sixteen.7 1000 from domicile plate so the brawl is moving horizontally, how long does it take the ball to reach habitation plate? (b) How far does the ball drop between the bullpen's hand and dwelling house plate?

A projectile is launched at an angle of [latex] 30\text{°} [/latex] and lands 20 southward later at the same height equally it was launched. (a) What is the initial speed of the projectile? (b) What is the maximum altitude? (c) What is the range? (d) Calculate the displacement from the point of launch to the position on its trajectory at 15 south.

A basketball role player shoots toward a handbasket 6.1 thou abroad and iii.0 chiliad above the floor. If the ball is released 1.8 m above the floor at an angle of [latex] threescore\text{°} [/latex] higher up the horizontal, what must the initial speed be if it were to go through the handbasket?

At a particular instant, a hot air balloon is 100 m in the air and descending at a constant speed of 2.0 m/south. At this verbal instant, a girl throws a ball horizontally, relative to herself, with an initial speed of 20 yard/south. When she lands, where will she find the ball? Ignore air resistance.

[latex] -100\,\text{one thousand}=(-2.0\,\text{one thousand}\text{/}\text{s})t-(4.ix\,\text{chiliad}\text{/}{\text{s}}^{2}){t}^{ii}, [/latex][latex] t=4.three\,\text{s,} [/latex][latex] x=86.0\,\text{m} [/latex]

A man on a motorcycle traveling at a uniform speed of 10 grand/s throws an empty can directly upwards relative to himself with an initial speed of 3.0 m/southward. Find the equation of the trajectory equally seen by a police officer on the side of the route. Presume the initial position of the tin is the indicate where it is thrown. Ignore air resistance.

An athlete can jump a altitude of eight.0 m in the broad bound. What is the maximum distance the athlete can jump on the Moon, where the gravitational acceleration is one-sixth that of Earth?

[latex] {R}_{Moon}=48\,\text{m} [/latex]

The maximum horizontal distance a male child can throw a brawl is 50 m. Presume he can throw with the same initial speed at all angles. How high does he throw the brawl when he throws it straight upwardly?

A stone is thrown off a cliff at an bending of [latex] 53\text{°} [/latex] with respect to the horizontal. The cliff is 100 g high. The initial speed of the stone is thirty m/s. (a) How high above the border of the cliff does the rock ascent? (b) How far has it moved horizontally when information technology is at maximum altitude? (c) How long afterwards the release does information technology striking the ground? (d) What is the range of the rock? (e) What are the horizontal and vertical positions of the rock relative to the border of the cliff at t = 2.0 s, t = 4.0 due south, and t = 6.0 s?

Trying to escape his pursuers, a hush-hush amanuensis skis off a slope inclined at [latex] 30\text{°} [/latex] below the horizontal at 60 km/h. To survive and land on the snow 100 m below, he must clear a gorge lx m broad. Does he brand it? Ignore air resistance.

A skier is moving with velocity v sub 0 down a slope that is inclined at 30 degrees to the horizontal. The skier is at the edge of a 60 m wide gap. The other side of the gap is 100 m lower.

A golfer on a fairway is 70 m away from the green, which sits beneath the level of the fairway by 20 m. If the golfer hits the brawl at an bending of [latex] 40\text{°} [/latex] with an initial speed of xx one thousand/s, how close to the green does she come?

A projectile is shot at a colina, the base of which is 300 m away. The projectile is shot at [latex] 60\text{°} [/latex] higher up the horizontal with an initial speed of 75 one thousand/south. The hill can be approximated by a plane sloped at [latex] xx\text{°} [/latex] to the horizontal. Relative to the coordinate organization shown in the post-obit effigy, the equation of this directly line is [latex] y=(\text{tan}twenty\text{°})x-109. [/latex] Where on the hill does the projectile land?

A projectile is shot from the origin at a hill, the base of which is 300 m away. The projectile is shot at 60 degrees above the horizontal with an initial speed of 75 m/s. The hill is sloped away from the origin at 20 degrees to the horizontal. The slope is expressed as the equation y equals (tan of 20 degrees) times x minus 109.

An astronaut on Mars kicks a soccer ball at an bending of [latex] 45\text{°} [/latex] with an initial velocity of 15 m/s. If the acceleration of gravity on Mars is 3.7 m/s, (a) what is the range of the soccer kick on a flat surface? (b) What would be the range of the same kick on the Moon, where gravity is one-sixth that of Earth?

Mike Powell holds the tape for the long jump of 8.95 chiliad, established in 1991. If he left the ground at an angle of [latex] 15\text{°}, [/latex] what was his initial speed?

MIT's robot chetah tin can jump over obstacles 46 cm loftier and has speed of 12.0 km/h. (a) If the robot launches itself at an angle of [latex] threescore\text{°} [/latex] at this speed, what is its maximum peak? (b) What would the launch angle have to be to reach a meridian of 46 cm?

Mt. Asama, Nihon, is an agile volcano. In 2009, an eruption threw solid volcanic rocks that landed 1 km horizontally from the crater. If the volcanic rocks were launched at an angle of [latex] twoscore\text{°} [/latex] with respect to the horizontal and landed 900 m below the crater, (a) what would be their initial velocity and (b) what is their time of flight?

Drew Brees of the New Orleans Saints can throw a football 23.0 g/s (l mph). If he angles the throw at [latex] 10\text{°} [/latex] from the horizontal, what distance does it go if it is to be caught at the same acme equally it was thrown?

[latex] R=18.5\,\text{chiliad} [/latex]

The Lunar Roving Vehicle used in NASA's late Apollo missions reached an unofficial lunar land speed of 5.0 m/s by astronaut Eugene Cernan. If the rover was moving at this speed on a flat lunar surface and hit a small bump that projected it off the surface at an angle of [latex] 20\text{°}, [/latex] how long would it be "airborne" on the Moon?

A soccer goal is 2.44 m high. A actor kicks the ball at a distance 10 m from the goal at an angle of [latex] 25\text{°}. [/latex] What is the initial speed of the soccer ball?

[latex] y=(\text{tan}\,{\theta }_{0})x-[\frac{g}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}]{x}^{2}⇒{v}_{0}=16.4\,\text{g}\text{/}\text{s} [/latex]

Olympus Mons on Mars is the largest volcano in the solar organisation, at a height of 25 km and with a radius of 312 km. If you lot are standing on the summit, with what initial velocity would you take to fire a projectile from a cannon horizontally to clear the volcano and land on the surface of Mars? Note that Mars has an acceleration of gravity of [latex] three.7\,\text{m}\text{/}{\text{southward}}^{2}. [/latex]

In 1999, Robbie Knievel was the first to bound the Thou Coulee on a motorbike. At a narrow office of the coulee (69.0 m broad) and traveling 35.8 m/s off the takeoff ramp, he reached the other side. What was his launch bending?

[latex] R=\frac{{five}_{0}^{2}\text{sin}\,two{\theta }_{0}}{g}⇒{\theta }_{0}=xv.0\text{°} [/latex]

You throw a baseball game at an initial speed of 15.0 yard/s at an angle of [latex] 30\text{°} [/latex] with respect to the horizontal. What would the ball's initial speed have to exist at [latex] 30\text{°} [/latex] on a planet that has twice the acceleration of gravity as Globe to achieve the same range? Consider launch and touch on a horizontal surface.

Aaron Rogers throws a football at 20.0 m/s to his wide receiver, who runs straight down the field at nine.4 grand/due south for xx.0 g. If Aaron throws the football when the wide receiver has reached x.0 m, what angle does Aaron have to launch the brawl and so the receiver catches it at the xx.0 1000 mark?

Glossary

projectile motion
motion of an object subject just to the dispatch of gravity
range
maximum horizontal distance a projectile travels
fourth dimension of flight
elapsed time a projectile is in the air
trajectory
path of a projectile through the air

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Source: https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/4-3-projectile-motion/

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