Question

A body released from a height falls freely towards the Earth. Another body is released from the same height exactly one second later. The separation between the two bodies two second after the release of the second body is
a) 9.8m
b) 49m
c) 24.5m
d) 19.6m

Answer 1

Hint: Both the bodies are falling toward the Earth and hence they both will accelerate due to gravity. The initial point of release is the same for both of them. It is given in the question that the first body is released one second before the second body and therefore two seconds after the release of the second body, the first body goes under acceleration for 3 seconds. From this information we can use Newton’s second kinematic equation to calculate the distance of separation. Take g as $9.8m{{s}^{-2}}$.

Complete step-by-step answer:

In the above figure we can see clearly that the time traveled body 1 is 3 sec and time traveled by body 2 is 2 sec. Assuming that both the bodies were at rest initially, Let us use a second kinematic equation to determine the distance travelled by both of them. Second kinematic equation is given by,
$S=Ut+\dfrac{1}{2}a{{t}^{2}}$ where S is the distance covered by the object in time t, U is the initial velocity of the object and a is the acceleration of the body.
In the above case the body is under acceleration due to gravity, and the initial velocities of both the bodies i.e. U=0. Now let us first calculate the distance covered by body 1, using the above equation.
$\begin{align}
  & S(\text{body 1})=Ut+\dfrac{1}{2}a{{t}^{2}}\text{, since U=0, a=9}\text{.8 and t=3 we get,} \\
 & S(\text{body 1})=\dfrac{1}{2}9.8{{(3)}^{2}}=44.1m \\
\end{align}$
Now let us calculate the distance covered by body 2 using the same equation.
$\begin{align}
  & S(\text{body 2})=Ut+\dfrac{1}{2}a{{t}^{2}}\text{, since U=0, a=9}\text{.8 and t=2 we get,} \\
 & S(\text{body 2})=\dfrac{1}{2}9.8{{(2)}^{2}}=19.6m \\
\end{align}$
Hence the distance of separation (d)between the two bodies is,
$\begin{align}
  & d=\text{distance covered by (body1)}-\text{distance covered by (body2)} \\
 & d=44.1-19.6=24.5m \\
\end{align}$

So, the correct answer is “Option c”.

Note: We have assumed that the initial velocities of both the bodies to be zero. It is clearly given in the question that the body is released from the same height and hence to be considered that both the bodies are at rest. The rate at which the velocity of the body does not depend on their individual masses.