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A ball is released from the top of a tower of height h meters. It takes T second to reach the ground. What is the position f the ball in T/3 second
initial velocity = u = 0
accleration = - g (downward)
height of tower = h
applying , S = Ut + at2/2
u = 0 , S = -h , a = -g , after putting these values
T = (2h/g)1/2 ..........1 (time taken by ball to reach the ground)
now again aplying , S = Ut + at2/2
u = 0 , a = -g , S = ? , t = T/3 , after putting these
S = gT2/18
put T from eq 1
S = h/9
therefore after T/3 sec , particle is H/9 distance from top of tower ...
approve if u like my ans
hi
its position will be 3h(rootg)-2(root h)/3(rootg) meters above the ground.............................
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