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.

Question

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.

drwls · Answer

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

Katie · Answer

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?