Question

A ball is released from the top of a tower of height h metres. It takes T sec. to reach the ground. what is the positive of the ball from ground in T/3 second

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Answer 1

so as we are throwing the ball from the tower, so

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

now,

time- t/3 sec

distance,s -?

again, using second eqn of motion

now putting eqn 1 here, we get

so ans is, distance travelled is h/9 metres from tower.

hope this helped

plss brainliest my answer.

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