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Question

A particle is dropped from the top of the tower. During its motion it covers 9/25 th part of the tower in the last one sec. Find the height of the tower.

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Solution

Given,

The height of tower =H

Total time is taken to reach the ground =T

16H25=0.5gT2(1)

9H25=0.5g(T1)2

Divide above two equations

169=(T(T1))2

43=T(T1)

3T=4T4

T=4s

Put the value ofT=4s in equation (1)

16H25=0.5g×42

H=(2516)0.5×9.8×16

H=122.5m

THus, Height of the tower is 122.5m


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