Question

An object is dropped from rest at a height of 150 m and simultaneously

another object is dropped from rest at a height 100 m. What is the difference

in their heights after 2 s if both the objects drop with same accelerations?

How does the difference in heights vary with time?

Answer 1

let

h₁ = 150 m and h₂ = 100 m

and we know

s = ut + (1/2)at²

or as in this case, u =0 and a = -g

so

s = -(1/2)gt²

or

s1 = s2 = -(1/2)gt² = -(1/2)x10x(2)² = 20 m

which is the distance travelled by the two objects.

now the height of the stone 1 after 2 seconds will be

h'₁ = 150 - 20 m = 130 m

and similarly for stone 2

h'₂ = 100 - 20 = 80 m

so, the difference in heights will still be = 130 - 80 m =50m

and their difference in height would stay constant until stone 2 touches the ground.