Question

A ball is released from a height 'h' reaches the ground in time 'T'. Where will it be from the ground at time 5T /6

Answer 1

As per the laws of motion

s=ut+0.5at^2

As ball is released,

∴u=0

s=0.5at^2

s=0.5gT^2 ------(1)

s'=0.5g(5T/6)^2 ------(2)

Divide both eq.

h/h'=36/25

h'=25h/36 (From roof)

∴From ground = h-25h/36

= 11h/36