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

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!!!

Approved

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.............................