Question

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

Answer 1

S = ut + 1/2 × at^2

U = 0

t = T/4

S = 0×T/4 + 1/2 × 10 × (T/4)^2

= 5 × T^2/16

Position of the body = h - 5T^2/16

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Answer 2

Answer:-

vT/2

Explanation:

since it is released from top from top of building so it is free fall

therefore u=0m/s

let final velocity with which it will strike the ground be v

t=T seconds

acceleration=(v-u)t

=>(v-0)T

=>v/T.........1

By using 2nd equation of motion

s=ut +1/2×a×t×t

=>0×T/4+1/2×v/T×T/4×T/4

=>s=vT/32

HENCE HIS POSITION AT T/4 SECONDS IS 'vT/32'

hope it is correct