Physics
posted by Lauryn .
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 = 6.70 m high, forming a 1.5mhigh railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of θ = 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 clears the wall.
I calculated this correctly and got 18.13 m/s.
(b) Find the vertical distance by which the ball clears the wall.
I know that dy=vy0*t but I tried using that equation and couldn't get the answer.
I did this and it was incorrect
18.13*sin53*2.2=dy
32.856.7=25.45
(c) Find the horizontal distance from the wall to the point on the roof where the ball lands.
I can't figure out how to do the last two, can anyone please help me set up these problems so I can figure out how to solve them? Thank you.

On part b,
You need to find how long it takes the ball to go th 24.0m in the x direction. Use that time for the y calculation. 
Sorry, I see that was given.

Yeah :( thanks for trying though, I tried using the v0sin53 and got 18.13*sin53= 14.48
then I tried plugging that into the y, to get y=14.48sin53(2.2)1/2(9.8)(2.2^2)
I got 25.4423.716=1.725, the computer assignment says that is incorrect. I don't know what I am doing wrong 
I calculated the ball was thrown at a speed of 18.12m/s.
Vx is 24m/2.2s=10.91m/s
Vx=V(cos(53))
That gives V as 18.12m/s 
I get 14.47m/s for Vy.

Using that...
Y=14.47m/s(2.2s)(1/2)(9.8m/s)(2.2)^2
Y=31.834m  23.716m
=8.118m
So it clears the wall by 8.118m  6.7m
=1.418m
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