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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`
6 years ago

Shivam Bhagat
22 Points
```										h=1/2*10*Tsqure
Tsqure=h/5...........1

f=1/2*10*Tsqure/9
From 1 .....
f=h/9

HEIGHT FROM GROUND=
h-h/9  =8h/9

All the best!!!
```
6 years ago
510 Points
```										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
```
6 years ago
iit jee
44 Points
```										hi
its position will be 3h(rootg)-2(root h)/3(rootg) meters above the ground.............................
```
6 years ago
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