A playground is on the flat roof of a city school, 4.9 m above the street below

Question

A playground is on the flat roof of a city school, 4.9 m above the street below (see figure). The vertical wall of the building is h = 6.40 m high, forming 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 = 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) let the speed be u, horizontal speed= ucos 53o=0.60u

0.60u*t=d

0.6u*2.2=24

u=24/(0.6*2.2)= 18.18m/s Answer

b) Lets find height of ball at this value time,

H=usin53 * t - 0.5gt2

H=0.8*18.18*2.2-0.5*9.8*2.2*2.2= 8.28 m

It clears the wall by 8.28-6.40= 1.88m

c) lets T be the total time taken to hit the roof

s=usin53 * T - 0.5gT2

4.9=0.8*18.18 T - 0.5*9.8T2

4.9T2 - 14.54T +4.9=0

values of T= 2.58, 0.39.....We will take larger value of T as it should be more than 2.20s

Extra time after wall= 2.58-2.20= 0.38 s

horizontal distance of striking point from the wall= 0.38 * ucos53o= 4.145 m =4.15 m

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