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Question

A body falls freely from the top of a tower. It covers 36% of the total height in the last second before striking the ground level. The height of the tower is

A
50 m
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B
75 m
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C
100 m
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D
125 m
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Solution

The correct option is D 125 m
Let the height of tower be h and the body takes time t to reach the ground when it falls freely.
Using second equation of motion, we have
h=ut+12gt2 (Take downward direction as positive)
h=12gt2 ( u=0 during free fall) .…(i)
In the last second, i.e., in tth second the body travels 0.36h. Thus, in |t1| seconds, it travels h0.36h=0.64h
So, we get, 0.64h=12g(t1)2 ....(ii)
On solving equ. (i) and (ii), we get
h0.64h=12gt212g(t1)2t=5 s
Substituting in equation (i), h=125 m

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