Question #65d5d

2 Answers
Jul 15, 2016

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)#

Jul 15, 2016

Answer:

#"Ratios is shown in figure"#

Explanation:

enter image source here

#"example: ratio for t=20 and t=21"#

#((2*20-1)*x)/((2*21-1)*x)=(39*x)/(41*x)=39/41#