A playground is on the flat roof of a city school, hb = 6.20 m above the street below (see figure). The vertical wall of the building is h = 7.70 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 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 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.

Without any concern for my personal welfare or credibility, I shall ASSUME that the point directly above the vertical wall that the ball reaches 2.2 sec after the kick is the HIGHEST point of its arc.

I shall ASSUME further that the force of the kick, and the subsequent trajectory of the ball, are in the direction perpendicular to the wall.

a). As soon as the ball is kicked, the vertical component of its velocity begins to decrese at the rate of 9.8m/s² ... the downward acceleration of gravity. It totally runs out of steam after 2.2 seconds, when its initial vertical component has decreased by (9.8 x 2.2) = 21.56 m/s . So the initial value of its vertical component was 21.56 m/s. The vertical component of launch velocity was 21.56 = |V| sin(53). |V| = 21.56 / sin(53) = 27 m/s .

b). The AVERAGE vertical speed of the ball , from the kick until the highest point of the arc, is 1/2(21.56 - 0) = 10.78 m/s . That time interval lasts 2.2 sec. So during that time, the ball rises (10.78 x 2.2) = 23.72 meters from the ground.

The highest point of the wall is 7.7 meters above ground, so the ball clears the top of the wall by (23.72 - 7.7) = 16 meters. (A healthy kick indeed.)

c). The ball begins to fall from 16 meters, down to the flat part of the roof at 6.2 meters ... a fall of (16 - 6.2) = 9.8 meters from the top of the arc. The time it takes to fall comes out of the 'drop' formula ... D = 1/2 g T² .

9.8 = 1/2 (9.8) T²

T = √ (9.8 / 4.9) = √2 seconds

During that time, the ball continues moving horizontally past the wall, over the roof, at the same horizontal velocity it has had since it was kicked. The horizontal component of velocity can be found from the initial conditions immediately after the big bang ...

tan(53) = 21.56 / H

H = 21.56 / tan(53) = 16.25 m/s

After cruising over the highest point of the wall, the ball continues sailing horizontally, at 16.25m/s, for the √2 seconds it takes to fall from the highest point of the arc to the roof. The ball lands

D = (speed) x (time) = (16.25 m/s) x (√2 sec) = 22.98 meters from the wall.

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Thank you for the 5 points. The dry crust and warm cloudy water are delicious,
with an earthy bouquet and a woody finish.