A playground is on the flat roof of a city school, 5.8 m above the street below

Question

A playground is on the flat roof of a city school, 5.8 m above the street below (see figure). The vertical wall of the building is h = 6.90 m high, forming 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. (Give your answer to two decimal places to reduce rounding errors in later parts.) 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

Explanation / Answer

a.) To find the speed at which the ball was launched:

D = v * t
24 m = (V*cos(53 degrees)) * 2.2s Vcos(angle) = horizontal speed
V = 18.12 m/s

b.) At that initial speed, the initial vertical component is:

Vy = V*sin(angle)
Vy = 18.12*sin(53 degrees)
Vy = 14.47 m/s

To get vertical distance, use:

d = Vy *t + (1/2) * a * t^2
d = 14.47 * 2.2+ (1/2) * (-9.8) * 2.2^2
d = 8.11m

So the ball clears by 8.11m - 6.9 m = 1.21m

c.) To find the distance past the wall, you need the time until it lands:

d = Vy * t + (1/2) * a * t^2
-8.67= 0 * t + (1/2) * -9.8 * t^2

t = 1.94 s

Then multiply the time by the horizontal speed to get the distance past the wall:

d = V * t
d = 18.12*cos(53 degrees) * 1.94 horizontal speed is the same as initially
d = 21.15m

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