A playground is on the flat roof of a city school, h b = 5.20 m above the street

Question

A playground is on the flat roof of a city school, hb = 5.20 m above the street below (see figure). The vertical wall of the building is h = 6.30 m high, to form a 1.1-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° 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.

(b) Find the vertical distance by which the ball clears the wall.

(c) Find the horizontal distance from the wall to the point on the roof where the ball lands.

Explanation / Answer

(a) Here, We have, 2.2cos(53)Vo = 24

=> Vo = 18.1 m/s

Now,
h = 2.2sin(53)(18.1) - 0.5*9.8*(2.2)^2 = 31.80 - 23.72 = 8.08 m
Assuming the passerby has an initial height of 0.
The ball is 8 meters up, which means it is 8 - 7 = 1 meter above the wall.
b) So, the vertical distance by which the ball clears the wall = 1 meter

New equation is now
5.2 = sin(53)(t)(18.1) - 4.9(t)^2

=> 5.2 = 14.45t - 4.9t^2

=> 4.9t^2 - 14.45t + 5.2 = 0

here, a = 4.9, b = -14.45, c = 5.2

t = [14.45 + sqrt(208.80 - 101.92)]/9.8 = 2.53 s.
There will be two times, you want the larger time. This time is 2.53 seconds which you will plug in to the first eq
cos(53)18.1*2.53 = d
d = 27.56 m

Now, d - 24 = 3.56 m
c) Hence, the horizontal distance from the wall to the point on the roof where the ball lands = 3.56 m.

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