# 3.3 Projectile motion  (Page 9/16)

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Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle $\theta$ below the horizontal. The service line is 11.9 m from the net, which is 0.91 m high. What is the angle $\theta$ such that the ball just crosses the net? Will the ball land in the service box, whose out line is 6.40 m from the net?

$\theta =6.1º$

yes, the ball lands at 5.3 m from the net

A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. (a) If the ball is thrown at an angle of $\text{25º}$ relative to the ground and is caught at the same height as it is released, what is its initial speed relative to the ground? (b) How long does it take to get to the receiver? (c) What is its maximum height above its point of release?

Gun sights are adjusted to aim high to compensate for the effect of gravity, effectively making the gun accurate only for a specific range. (a) If a gun is sighted to hit targets that are at the same height as the gun and 100.0 m away, how low will the bullet hit if aimed directly at a target 150.0 m away? The muzzle velocity of the bullet is 275 m/s. (b) Discuss qualitatively how a larger muzzle velocity would affect this problem and what would be the effect of air resistance.

(a) −0.486 m

(b) The larger the muzzle velocity, the smaller the deviation in the vertical direction, because the time of flight would be smaller. Air resistance would have the effect of decreasing the time of flight, therefore increasing the vertical deviation.

An eagle is flying horizontally at a speed of 3.00 m/s when the fish in her talons wiggles loose and falls into the lake 5.00 m below. Calculate the velocity of the fish relative to the water when it hits the water.

An owl is carrying a mouse to the chicks in its nest. Its position at that time is 4.00 m west and 12.0 m above the center of the 30.0 cm diameter nest. The owl is flying east at 3.50 m/s at an angle $30.0º$ below the horizontal when it accidentally drops the mouse. Is the owl lucky enough to have the mouse hit the nest? To answer this question, calculate the horizontal position of the mouse when it has fallen 12.0 m.

4.23 m. No, the owl is not lucky; he misses the nest.

Suppose a soccer player kicks the ball from a distance 30 m toward the goal. Find the initial speed of the ball if it just passes over the goal, 2.4 m above the ground, given the initial direction to be $\text{40º}$ above the horizontal.

Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s.

No, the maximum range (neglecting air resistance) is about 92 m.

The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 7.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved in solving projectile motion problems.

In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77 m. What was the initial speed of the shot if he released it at a height of 2.10 m and threw it at an angle of $38.0º$ above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at $\text{45º}$ when air resistance is neglected, the actual angle to achieve maximum range is smaller; thus, $\text{38º}$ will give a longer range than $\text{45º}$ in the shot put.)

15.0 m/s

A basketball player is running at directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity. (a) What vertical velocity does he need to rise 0.750 m above the floor? (b) How far from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket?

A football player punts the ball at a $45.0º$ angle. Without an effect from the wind, the ball would travel 60.0 m horizontally. (a) What is the initial speed of the ball? (b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.50 m/s. What distance does the ball travel horizontally?

(a) 24.2 m/s

(b) The ball travels a total of 57.4 m with the brief gust of wind.

Prove that the trajectory of a projectile is parabolic, having the form $y=\text{ax}+{\text{bx}}^{2}$ . To obtain this expression, solve the equation $x={v}_{0x}t$ for $t$ and substitute it into the expression for $y={v}_{0y}t–\left(1/2\right){\text{gt}}^{2}$ (These equations describe the $x$ and $y$ positions of a projectile that starts at the origin.) You should obtain an equation of the form $y=\text{ax}+{\text{bx}}^{2}$ where $a$ and $b$ are constants.

Derive $R=\frac{{v}_{0}^{2}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{2\theta }_{0}}{g}$ for the range of a projectile on level ground by finding the time $t$ at which $y$ becomes zero and substituting this value of $t$ into the expression for $x-{x}_{0}$ , noting that $R=x-{x}_{0}$

$y-{y}_{0}=0={v}_{0y}t-\frac{1}{2}{\mathrm{gt}}^{2}=\left({v}_{0}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \right)t-\frac{1}{2}{\mathrm{gt}}^{2}$ ,

so that $t=\frac{2\left({v}_{0}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \right)}{g}$

$x-{x}_{0}={v}_{0x}t=\left({v}_{0}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \right)t=R,$ and substituting for $t$ gives:

$R={v}_{0}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \left(\frac{{2v}_{0}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta }{g}\right)=\frac{{2v}_{0}^{2}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta }{g}$

since $2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =\text{sin}\phantom{\rule{0.25em}{0ex}}2\theta ,$ the range is:

$R=\frac{{{v}_{0}}^{2}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}2\theta }{g}$ .

Unreasonable Results (a) Find the maximum range of a super cannon that has a muzzle velocity of 4.0 km/s. (b) What is unreasonable about the range you found? (c) Is the premise unreasonable or is the available equation inapplicable? Explain your answer. (d) If such a muzzle velocity could be obtained, discuss the effects of air resistance, thinning air with altitude, and the curvature of the Earth on the range of the super cannon.

Construct Your Own Problem Consider a ball tossed over a fence. Construct a problem in which you calculate the ball’s needed initial velocity to just clear the fence. Among the things to determine are; the height of the fence, the distance to the fence from the point of release of the ball, and the height at which the ball is released. You should also consider whether it is possible to choose the initial speed for the ball and just calculate the angle at which it is thrown. Also examine the possibility of multiple solutions given the distances and heights you have chosen.

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