1.

A ball is released from the top of a tower of height h meter . It takes t second to reach the ground . What is the position of ball at t/3 second?

Answer»

We have, initial velocity, u = 0

Acceleration, G = 9.8 m/s2

Now,

S = ut + ½ at2

=> H = 0 + ½ (9.8)T2

=> h = ½ (9.8)T2

Again,

S = ut + ½ at2

=> S = 0 + ½ (9.8)(T/3)2

=> S = (1/9)[½(9.8)T2]

=> S = h/9

Thus, after T/3 SECONDS the ball is at a DISTANCE h/9 meters from the top. Or, it is (h – h/9 =) 8h/9 meters from the ground.


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