# college Physics I

posted by Katie

A playground is on the flat roof of a city school, 5.2 m above the street below (see figure). 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 è = 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.m/s
b) Find the vertical distance by which the ball clears the wall. m
c Find the horizontal distance from the wall to the point on the roof where the ball lands. m

I forgot to add the questions to my question earlier. Here they are now. Thanks for your help.

1. drwls

a) The horizontal component of the launch speed (Vo) is Vo cos = 0.6018 V. It travels 24 m horizontally in 2.20 s. Therefore
0.618 Vo * 2.2 = 24.0. Solve for Vo
Vo = 17.65 m/s

The vertical lauch speed component is Vo sin 53 = 14.10 m/s. Use the equation of vertical motion,

y = (Vo sin 53)* t - (1/2) g t^2

to solve for the height after t = 2.2 s

c) Use the same equation to solve for t when y = 5.2 m

2. Katie

I have tried to put the numbers in the equation, but I am not getting the correct answer. Is there an easier way to explain it?

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