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Question

Ball A is dropped from a 44.1m high cliff. Two seconds later, another Ball B is thrown downwards from the same place with some initial speed. The two balls reached the ground together. Find the speed with which the ball B was thrown.

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Solution

time take by first ball to reach the ground

use eqn of motion s=ut+at2/2,u=0,s=44.1m,a=9.8m/s2

44.1=9.8t2/2

t=3sec

time taken by second ball =3-2=1sec

now we find initial velocity for second ball

use eqn of motion s=ut+at2/2

44.1=u(1)+9.8(1)2/2

u=39.3m/s


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