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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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
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