A playground is on the flat roof of a city school, 5.2 m above the street below

Question

A playground is on the flat roof of a city school, 5.2 m above the street below . The vertical wall of the building is h = 6.50 m high, forming a 1.3-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 = 53.0° above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall.

(c) Find the horizontal distance from the wall to the point on the roof where the ball lands.

Explanation / Answer

Horizontal component of ball's velocity is
24 m/2.20 s = 10.91 m/s

So the ball's initial speed was
10.91/ cos(53) = 18.127 m/s

y = 18.127 * t *sin(53) - (1/2) g t^2
At t = 2.20 s,
y = 14.477 * 2.2 - 4.9*2.2^2
= 8.13 m
so the ball clears the 6.5-meter wall by 1.63 m

ball lands when y = 5.2 m on the way down.
5.2 = 14.477 t - (4.9) t^2
0 = 4.9t^2 - 14.477 t + 5.2
t>2.2 because it is in way down
t = 2.536 seconds
Total x = (10.91 m/s) (2.536 s) = 27.66 m
This is 24 + 3.66, so the distance from the outer surface of the wall
to the point where the ball lands on the roof is 3.66 meters.

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