Question

39. A body released from the top of a tower of height h takes T seconds to reach the ground. Th
position of the body at T /4 seconds is​

Answer 1

Answer:

H/16 from top of tower

Explanation:

From kinematic equation

S=ut+1/2at^2

H=(0)T+1/2gT^2

H=1/2gT^2 .......eq1

Let position of body be H°

H°= uT/4+1/2g(T/4)^2

H°=1/2g×T^2/16

H°=H/16 .....from eq1

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