College Physics
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 1.3mhigh 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 posted this question earlier and I didn't understand the feedback. Could someone explain it differently? I need it step by step to understand it. Thanks!

I provided the "feedback", in steps. What didn't you understand about it?

For the speed at which the ball was launched I got 18.12m/s. That part was right according to the computer. I didn't understand parts b or c of the question. For part a you multiplied vt by .608 instead of .6018 and that is why I got a different answer. Could you explain parts b and c again? Thanks.

a) 18.12
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