Question

a body dropped from a height h reaches the ground at a time t.find the time taken to reach a height h/2​

Answer 1

Explanation:

Total Height=h

Time taken to cover this total height=t

Let Height after t/2 seconds be H and let t/2 =T.

By Equation of Motion,

h=ut+1/2gt^2 (downward direction taken positive)

u=0 (body is dropped)

therefore h=1/2gt^2———————(EQ.1)

now,

H=1/2gT^2

H=1/2g(t/2)^2

H=(1/2gt^2)/4

but we know for EQ.1 that 1/2gt^2=h

therefore, H=h/4

ANS- H=h/4

Answer 2

Answer:

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Explanation:

Total Height=h

Time taken to cover this total height=t

Let Height after t/2 seconds be H and let t/2 =T.

By Equation of Motion,

h=ut+1/2gt^2 (downward direction taken positive)

u=0 (body is dropped)

therefore h=1/2gt^2———————(EQ.1)

now,

H=1/2gT^2

H=1/2g(t/2)^2

H=(1/2gt^2)/4

but we know for EQ.1 that 1/2gt^2=h

therefore, H=h/4

ANS- H=h/4