Question

The ratio of distances travelled by a freely falling body is 1 sec,2 sec,3 sec is ????

Answer 1

1:2...................

Answer 2

we know that S = ut + .5 at^2
but for a freely falling object
u=0 a = g (9.81 m/s^2) s = h ( distance )
so equation becomes
h = .5 gt^2
now substitute 1 sec , 2 sec and 3 sec.... so
h1 = .5*9.81*1^2
h2 = constant* 2^2
h3= constant* 3^2
so h1:h2:h3:: 1: 4:9