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Grade: 12
        

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

8 years ago

Answers : (3)

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!!!
						
8 years ago
vikas askiitian expert
509 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

8 years ago
iit jee
44 Points
							

hi

its position will be 3h(rootg)-2(root h)/3(rootg) meters above the ground.............................

8 years ago
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