# 0.4 2.5 projectile motion  (Page 6/13)

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

where ${v}_{0}$ is the initial speed and ${\theta }_{0}$ is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described.

## Summary

• Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity.
• To solve projectile motion problems, perform the following steps:
1. Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position $\mathbf{s}$ are given by the quantities $x$ and $y$ , and the components of the velocity $\mathbf{v}$ are given by ${v}_{x}=v\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta$ and ${v}_{y}=v\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ , where $v$ is the magnitude of the velocity and $\theta$ is its direction.
2. Analyze the motion of the projectile in the horizontal direction using the following equations:
$\text{Horizontal motion}\left({a}_{x}=0\right)$
$x={x}_{0}+{v}_{x}t$
${v}_{x}={v}_{0x}={\mathbf{\text{v}}}_{\text{x}}=\text{velocity is a constant.}$
3. Analyze the motion of the projectile in the vertical direction using the following equations:
$\text{Vertical motion}\left(\text{Assuming positive direction is up;}\phantom{\rule{0.25em}{0ex}}{a}_{y}=-g=-9\text{.}\text{80 m}{\text{/s}}^{2}\right)$
$y={y}_{0}+\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t$
${v}_{y}={v}_{0y}-\text{gt}$
$y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\text{gt}}^{2}$
${v}_{y}^{2}={v}_{0y}^{2}-2g\left(y-{y}_{0}\right).$
4. Recombine the horizontal and vertical components of location and/or velocity using the following equations:
$s=\sqrt{{x}^{2}+{y}^{2}}$
$\theta ={\text{tan}}^{-1}\left(y/x\right)$
$v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}$
${\theta }_{\text{v}}={\text{tan}}^{-1}\left({v}_{y}/{v}_{x}\right).$
• The maximum height $h$ of a projectile launched with initial vertical velocity ${v}_{0y}$ is given by
$h=\frac{{v}_{0y}^{2}}{2g}.$
• The maximum horizontal distance traveled by a projectile is called the range . The range $R$ of a projectile on level ground launched at an angle ${\theta }_{0}$ above the horizontal with initial speed ${v}_{0}$ is given by
$R=\frac{{v}_{0}^{2}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{2\theta }_{0}}{g}.$

## Conceptual questions

Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither $\text{0º}$ nor $\text{90º}$ ): (a) Is the velocity ever zero? (b) When is the velocity a minimum? A maximum? (c) Can the velocity ever be the same as the initial velocity at a time other than at $t=0$ ? (d) Can the speed ever be the same as the initial speed at a time other than at $t=0$ ?

Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither $\text{0º}$ nor $\text{90º}$ ): (a) Is the acceleration ever zero? (b) Is the acceleration ever in the same direction as a component of velocity? (c) Is the acceleration ever opposite in direction to a component of velocity?

## Problems&Exercises

A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of $30.0º$ above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the $x$ and $y$ distances from where the projectile was launched to where it lands?

$\begin{array}{lll}x& =& \text{1.30 m}×{10}^{2}\\ y& =& \text{30}\text{.9 m.}\end{array}$

A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

(a) 3.50 s

(b) 28.6 m/s (c) 34.3 m/s

(d) 44.7 m/s, $50.2º$ below horizontal

The cannon on a battleship can fire a shell a maximum distance of 32.0 km. (a) Calculate the initial velocity of the shell. (b) What maximum height does it reach? (c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is $6\text{.}\text{37}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{km}$ . How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?

(a) 560 m/s

(b) $8\text{.}\text{00}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m}$

(c) 80.0 m. This error is not significant because it is only 1% of the answer in part (b).

An arrow is shot from a height of 1.5 m toward a cliff of height $H$ . It is shot with a velocity of 30 m/s at an angle of $\text{60º}$ above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow’s impact speed just before hitting the cliff?

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

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.

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.

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.

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An helicopter is flying over new York with a horizontal component of velocity of 14.6m/s-1 and a vertical component of -8.62 m/s-1, calculate, (a), the magnitude of the total velocity of the helicopter. (b), the angle of the total velocity.