16.The ratio of distance travelled by a freely falling body 2nd s and 3rd s is

solution:

For a freely falling body initial velocity (u)=0

for a body which travelled in nth second distance

^{Sn=u+ a(n-1/2)}

for freely falling body acceleration(a)=g

for 2nd second S₂=g(2-1/2)

=g(3/2)

for 3rd second S₃=g(3-1/2)

=g(5/2)

the ratio between is 3/2g:5/2g=3:5