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 = 6.20 m high, to form a 1.5-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 horizontal distance from the wall to the point on the roof where the ball lands.

Explanation / Answer

Since the returned ball takes 2.2s. to its max. height, it reaches a height of 1/2 (t^2 x g) = 23.71 metres.
Initial vertical V component = (gt) = 9.8 x 2.2 = 21.56m/sec.
The ball also takes 2.2s. to reach the wall, so horizontal V component = (24/2.2) = 10.91m/sec.
Launch speed = sqrt.(10.91^2 + 21.56^2) = 24.16m/sec.
(23.71 - 6.20) = 17.51 metres.
The ball has to drop vertically (17.51 + 1.5) = 19.01 metres to the roof.
Time to drop = sqrt.(2h/g) = 1.26 secs.
In that time, it has advanced (1.26 x 10.91) = 13.80 metres horizontally, from the wall.

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