Question

A body is released from the top of the tower of heights h metre. It takes time T second to reach the ground. Where is the body at time T\2 seconds

Answer 1

Answer:

initial velocity,u= 0 m/s

time taken = t sec

distance travelled = h metres

acceleration( gravitational force) = 9.8 m s^- 2

using second eqn of motion

s = ut + at^2 / 2

h= 0 *t + 9.8*t^2/2

Explanation:

Answer 2

Explanation:initial velocity,u= 0 m/s

time taken = t sec

distance travelled = h metres

acceleration( gravitational force) = 9.8 m s^- 2

using second eqn of motion

s = ut + at^2 / 2

h= 0 *t + 9.8*t^2/2

hopes it is okk