A playground is on the flat roof of a city school, 5 m above the street below (s

Question

A playground is on the flat roof of a city school, 5 m above the street below (see figure). The vertical wall of the building is h = 6.20 m high, forming a 1.2-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 theta = 53.0 degree 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. Find the speed at which the ball was launched. (Give your answer to two decimal places to reduce rounding errors in later parts.) Find the vertical distance by which the ball clears the wall. Find the horizontal distance from the wall to the point on the roof where the ball lands.

Explanation / Answer

(a) Find the speed at which the ball was launched.
Let the speed of the ball at the launch 24 m away from the wall on the street be v.
We have 2.20*(v cos 53) = 24
From first equation we get v = 24/[2.20*cos 53] =18.12 m/s (to three significant figures)

(b) Find the vertical distance by which the ball clears the wall. the vertical component of the velocity, v' of ball when it reaches above the point vertically above the top of the wall is given by
v' = v sin53 - 9.8*2.2 = -5.88
vertical distance, H travelled is given by
H = [(v sin 53)^2 - v'^2]/(2*9.8)= 10.7 m. So the height h which is above the wall is 10.7- 6.20 = 4 .5 m

From second equation we get h by substituting for v as
h = [(18.12*sin 53)/2]*2.20 - 6.20 =9.71 m

(c) Find the distance from the wall to the point on the roof where the ball lands.
Time, T required to reach the maximum height = [18.12* sin 53]/9.8 =1.476 s
Maximum height reached will be = [18.12* sin 53]^2/(2*9.8 ) = 10.68 m
Time, T' required to fall through 10.68 - 5 =5.68m is given by
T' = sq rt [(2*5.68)/9.8] = 1.076 s
The horizontal distance, R' travelled from the instant the ball was launched to when it fell on playground is given by
R' = [18.12*cos 53]*(T + T') = 50.20 m
So the distance from the wall = 0.20 - 24 =26.2m

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