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25P

Expert-verifiedFound in: Page 150

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.**

- The ball was launched with a velocity of $18.127m/s$
- The ball clears the wall by 1.133m
- Horizontal distance from the wall to the point on the roof where the ball lands is 2.79m

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

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

Where,

$\theta =53\xb0$

When the ball is directly above the wall,

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

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

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

$u=18.127m/s$

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

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

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

$y=31.849-23.716$

$y=8.133m$

Since the height of the wall is 7m

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

For first half of the trajectory,

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

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

$s=10.69m$

For second half, solving for the final velocity,

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

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

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

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

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

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

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

$-9.59=14.477-9.8t$

$t=2.456s$

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

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

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

$x=26.79m$

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

**$26.79-24=2.79m$**

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