Question

two bodies of different masses m1 and m2 are dropped from two different heights "a" and "b"

what is the ratio of time taken by the two bodies to drop through these distances

Answer 1

we shall use the following kinematic relation

s = ut + (1/2)at²

here,

u = 0 and a = g, so

s = (1/2)gt²

now for object 1

s₁ = (1/2)gt₁²

or

t₁² = 2ag

thus,

t₁ = (2gs₁)^1/2

and

for object 2

s₂ = (1/2)gt₂²

thus,

t₂ = (2gs₂)^1/2

so, the ratio of their times will be

t1/t2 = (2gs₁)^1/2 / (2gs₂)^1/2

or

t₁ / t₂ = s₁^1/2 / s₂^1/2