Question

A playground is on the flat roof of a city school, h_b = 5.30 m above the street below. The vertical wall of the building is h = 6.80 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

Answer 1

  Horizontal component of ball's velocity is
24 m/2.20 s = 10.91 m/s

So the ball's initial speed was
10.91 m/s divided by cos(53 degrees) = 18.127 m/s

y = 18.127 m/s t sin(53 degrees) - (1/2) g t^2
At t = 2.20 s,
y = 14.477 m times 2.2 - (4.9 m/s^2) (2.2 s)^2
= 8.133 m
so the ball clears the 6.7-meter wall by 1.43 m

ball lands when y = 5.3 m on the way down.
5.3 = 14.477 t - (4.9) t^2
0 = 4.9t^2 - 14.477 t + 5.3
t = (14.477 plus or minus sqrt(14.477^2 - 4*5.3*4.9)) / 9.8
We will use the "plus" because the "minus" was on the way UP.
t = (14.477 plus sqrt(105.7)) /9.8
= 2.52 seconds
Total x = (10.91 m/s) (2.52 s) = 27.5 m
This is 24 + 3.5, so the distance from the outer surface of the wall
to the point where the ball lands on the roof is 3.5 meters