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A ball is projected horizontally with a velocity of20ms-1 from the top of the tower of height 78.4m. at the same time a boy who is 140m away from the foot of the tower starts running to catch the ball. With what constant velocity he must run to catch the ball before hitting the ground (neglect the height of the boy)

Asked by valavanvino1011 12th August 2018, 6:02 PM
Answered by Expert
The ball has to travel a vertical distance 78.4 m.
Hence the time to reach the ground is calaculated using the formula "S = (1/2)gt2 " because intial vertical component of velocity is zero.
Hence the time t to reach the ground is given by,  78.4 = (1/2)×9.8×t2  ...............(1)
solving the eqn.(1) for t, we get t = 4 s
in 4 seconds, the ball has moved horizontal distance 20×4 = 80 m. 
Since the boy initially at a distance 140 m from tower, he has to run a distance 60 m in 4 s.
Hence his speed 60/4 = 15 m/s

Answered by Expert 13th August 2018, 12:02 PM
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