Question

a body is released from the top of a tower at Nita takes t seconds to reach the ground where is the body at the time t by 2 seconds​

Answer 1

Answer:

Explanation:

Since the body is dropped,

u=0 H is negative

According to second eqn of motion,

S=ut +1/2at^2

-H= 0*t- 1/2gt^2

H=1/2 gt^2 ___1

At t=t/2

H'= 1/2g(t/2)^2

H'=1/2gt^2/4

From 1,

H'=1/4 H

Distance above ground =H-H/4

=3H/4