A playground is on the flat roof of a city school, 4.7 m above the street below

Question

A playground is on the flat roof of a city school, 4.7 m above the street below (see figure). The vertical wall of the building is h = 6.10 m high, forming a 1.4-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. (a) Find the speed at which the ball was launched. (Give your answer to two decimal places to reduce rounding errors in later parts.) m/s (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.

Explanation / Answer

(A) In horizontal,

d = (v0x) t

24 = (v0 cos53) (2.20)

v0 = 18.1 m/s .......Ans

(B) yf - yi = v0y t + ay t^2 / 2

H - 0 = (18.1 sin53)(2.20) - (9.8 x 2.20^2 / 2)

H = 8.09 m

delta(h) = H - h = 1.99 m ........Ans

(C) 4.7 - 0 = (18.1 sin53) t - 9.8 t^2 / 2

4.9 t^2 - 14.5 t + 4.7 = 0

t = 2.59 sec

x = (18.1 cos53) (2.59)

x = 28.2 m

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