Kinematic equation for a free falling body under gravity is given by
s=ut+1/2g t^2
Let s_1 be the distance traveled in time t_1
similarly s_2 and s_3 be distances traveled in time t_2 and t_3 respectively. As per original conditions
s_1=1/2g t_1^2
s_2=1/2g t_2^2
s_3=1/2g t_3^2
Distance traveled in time interval t_2-t_1 =s_2-s_1
=>s_2-s_1=1/2g t_2^2-1/2g t_1^2
=>s_2-s_1=1/2g( t_2^2- t_1^2) ......(1)
Similarly s_3-s_2=1/2g( t_3^2- t_2^2) ......(2)
Ratio of the distances traveled in successive intervals can be found by dividing (1) by (2)
(s_2-s_1)/(s_3-s_2)=(1/2g( t_2^2- t_1^2))/(1/2g( t_3^2- t_2^2))
(s_2-s_1)/(s_3-s_2)=( t_2^2- t_1^2)/( t_3^2- t_2^2)
