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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 = 7.30 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.

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.

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1 Answer

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a) The horizontal component of the launch speed (Vo) is Vo cos = 0.6018 V. It travels 24 m horizontally in 2.20 s. Therefore

X= X_0+V_x0(t)

24= 0+V_x(2.2)

10.91= V_0cos53

V_0= 18.13

b)to solve for the height after t = 2.2s

y = (Vo sin 53)* t - (1/2) g t^2

= ((18.13sin53)*2.2)- 4.9(2.2^2)

= 31.86-23.72

y = 8.14

8.14-7.3= .84

c) Use the same equation to solve for t when y = 5.8 m

5.8= ((18.13sin53)t)-4.9t^2

= 4.9t^2+14.18t+5.8

I suggest that you graph this on your calculator to solve for t

whatever your answer is for t plug it in to this equation:

x= V_x(t)

then subtract 24.0 m from your answer which will finally give u the answer for c

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