Question

An object is dropped from a height of 150 m at the same time another object is dropped from a
height of 100 m from the ground. What is the difference in their heights after 2 s from the ground​

Answer 1

Answer:

As the objects are dropped from rest their initial velocity is zero.

The acceleration will be acceleration due to gravity, which will be equal to 10 ms-2.

The height of the particle is written as:

D = H-S

Where H is initial height and S is displacement.

If we write heights of both the particles by using second equation of motion: s = ut + \frac{1}{2}at^2

D_{1}=H_{1}- \left ( ut + \frac{1}{2}at^{2} \right )

D_{2}=H_{2}- \left ( ut + \frac{1}{2}at^{2} \right )

Hence,

D_{1}-D_{2}=H_{1}-H_{2}=50m

The difference in height at any time is coming out to be 50 m.

So even, after two seconds, the difference will be 50 m and it will not vary with time.