Question

An object is dropped from rest at a height of $150m$ and simultaneously another object is dropped from rest at a height of $100m$. What is the difference in their heights after $2s$ if both the objects drop with same accelerations? How does the difference in heights vary with time?

• A. $30m$
• B. $40m$
• C. $50m$
• D. $60m$

Answer 1

Answer: (C)

Step 1: Given that:

The objects are dropped from rest that is

The initial velocity of both objects = $u_1=u_2=0$

Height from which the fist object is dropped= $H_1=150m$

Height from which the second object is dropped at the same time= $H_2=100m$

Acceleration ion first body= Acceleration in the other body

Given time($t$) = $2sec$
Step 2: Calculation of the distance covered by both the bodies in 2sec:

The distance covered by a body in downward direction under acceleration due to gravity that is $g$ is given by;

$h=ut+\frac{1}{2}g{t}^2$

For $u=0$

$h=0+\frac{1}{2}g{t}^2$

$h=\frac{1}{2}g{t}^2$

Now,

The distance travelled by the first body in $2s$ will be;

$h_1=\frac{1}{2}g\times {\left(2s\right)}^2$

$h_1=\frac{1}{2}g\times 4$

$h_1=2g$

The distance travelled by the other body in the same $2s$ will be given by;

$h_2=\frac{1}{2}g\times {\left(2s\right)}^2$

$h_2=\frac{1}{2}g\times 4$

$h_2=2g$

Step 3: Calculation of the difference in their heights after $2s$:

The difference in the heights of both the bodies initially is given by;

$=H_1-H_2$

$=150m-100m$

$=50m$

After $2s$,

The height at which the first body will be given by;

$=H_1-h_1$

$=150m-2g$

$=150-2g$

The height at which the other body will be given by;

${=H}_2-h_1$

$=100m-2g$

$=100-2g$

Now,

The difference in their heights will be given by:

$=(H_1-h_1)-(H_2-h_2)$

$=(150-2g)-(100-2g)$

$=150-2g-100+2g$

$=50m$

Which is the same as the difference in their heights initially.

Thus,

It is clear that the difference in their heights is $50m$ and it does not vary with time.

Thus,

Option C) $50m$ is the correct option.