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Found in: Page 150

### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

# A playground is on the flat roof of a city school, 6.00 m above the street below (Fig. P4.25). the vertical wall of the building is h 5 7.00 m high, forming a 1-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of u 5 53.0° above the horizontal at a point d 5 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall.(a) Find the speed at which the ball was launched.(b) Find the vertical distance by which the ball clears the wall.(c) Find the horizontal distance from the wall to the point on the roof where the ball lands.

1. The ball was launched with a velocity of $18.127m/s$
2. The ball clears the wall by 1.133m
3. Horizontal distance from the wall to the point on the roof where the ball lands is 2.79m
See the step by step solution

## Step 1: Taking the horizontal and vertical components of initial velocity

${u}_{x}=u\mathrm{cos}\theta$

${u}_{y}=u\mathrm{sin}\theta$

Where,

$\theta =53°$

## Step 2: Solving for initial velocity (a)

When the ball is directly above the wall,

$x={x}_{i}+{u}_{x}t+\frac{1}{2}{a}_{x}{t}^{2}$

$24=0+{u}_{x}\left(2.2\right)+0$

$u\mathrm{cos}53=\frac{24}{2.2}\left(Since,{u}_{x}=u\mathrm{cos}\theta \right)$

$u=18.127m/s$

Hence, the speed at which the ball was launched came out to be $18.127m/s$$18.127m/s$ .

## Step 3: Height from the ground when the ball passes over the wall (b)

$y={y}_{i}+{u}_{y}t+\frac{1}{2}{a}_{y}{t}^{2}\sum _{}^{}$

$y=0+18.127×\mathrm{sin}\left(53\right)×2.2+\frac{1}{2}\left(-9.8\right)2.{2}^{2}$

$y=31.849-23.716$

$y=8.133m$

Since the height of the wall is 7m

Therefore the ball clears the wall by $8.133-7=1.133m$

## Step 4: Now solving for the total distance in x-direction(c)

For first half of the trajectory,

${{v}_{y1}}^{2}-{{u}_{y1}}^{2}=2{a}_{y}s$

$0-\left[18.127\mathrm{sin}\left(53\right){\right]}^{2}=2\left(-9.8\right)s$

$s=10.69m$

For second half, solving for the final velocity,

${{v}_{y2}}^{2}-{{u}_{y2}}^{2}=2{a}_{y}\left[\left(h-1\right)-s\right]$

${{v}_{y2}}^{2}-0=2\left(-9.8\right)\left[\left(7-1\right)-10.69\right]$

${{v}_{y2}}^{2}=-91.924$

${v}_{y2}=-9.59m/s$

$Negativesigndenotesthatthevelocityisgoingdownwards.\phantom{\rule{0ex}{0ex}}Hence,thetotaltimetaken,$

${v}_{y2}={u}_{y}+{a}_{y}t$

$-9.59=18.127\mathrm{sin}\left(53\right)-9.8t$

$-9.59=14.477-9.8t$

$t=2.456s$

Hence, the distance travelled in time $t=2.546s$

$x={u}_{x}t$

$x=18.127\mathrm{cos}\left(53\right)×2.456$

$x=26.79m$

Therefore, the horizontal distance from the wall to the point of the roof where the ball lands is

$26.79-24=2.79m$