Question

A ball is released from the top of a tower of height h meters . It takes T seconds to reach the ground .What is the position of the ball in T/3 second? ?

Answer 1

s=ut + 1/2at^2

h=(0) + 1/2gt^2                          [u=0 as ball is dropped so it comes from rest]

h=1/2 gT^2-----------------------------(1)

distance travelled in T/3 seconds:

s=ut + 1/2at^2

s=(0) + 1/2g((T^2)/3^2)

s=1/18 g T^2

s=h/9                                             [from equation (1)}

THEREFORE BODY TRAVELS (h/9 ) DISTANCE IN TIME T/3

THEREFORE ITS POSITION FROM GROUND IS= h - h/9

                                                                              = 8h/9