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Question

A stone is released from a top. Another stone is dropped from a point 15 m below the top, when the first stone has reached a point 5 m below the top. Both stones reach the ground simultaneously. Determine the height of the top.

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Solution

When second ball is dropped, the two balls are 10 m apart. The velocity of first ball when it reaches 10 m below the top is given by,v2-u2=2ghv2-0=2×10×5or v=10 m/sSuppose the balls take time t to reach the ground. First ball travels 10 m more than second ball.So,10+(10t+12×10×t2)=12×10×t2or 10 t=10 or t=1 secondIn this time, the distance travelled by the second ball, y=12×10×t2=12×10×12=5 mBut, second ball is 15 m below the top. Thus, height of the top is 15+5=20m

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