Question

A ball is dropped downwards. After 2 seconds another ball is dropped downwards from the same point. What is the distance between them after 3 seconds?
(A) 25m
(B) 20m
(C) 50m
(D) 9.8m

Answer 1

Hint: In this problem, the body is being just dropped so it does not have any initial velocity. Also, no external force is acting on the system and the only force which is driving the motion of the ball towards the earth is the force of attraction due to gravity.

Complete step by step answer:
Let the first ball by ball A and the second ball be ball B. both of these balls are freely falling under the influence of gravity so the value of acceleration for both of them is g.
Let us take g= \[10m/{{s}^{2}}\]
FOR A;
Initial velocity, \[u=0\]
Acceleration, a=+g= \[10m/{{s}^{2}}\]
Distance covered by the ball A in 3 seconds can be calculated using Newton’s second equation of motion,
\[s=ut+\dfrac{a{{t}^{2}}}{2}\]
$ \implies {{s}_{1}}=\dfrac{10\times {{3}^{2}}}{2}=45m $
So the body A covers 45m
FOR B;
It was dropped after 2 seconds and time for which ball B travels is 3-1=2s
\[s=ut+\dfrac{a{{t}^{2}}}{2}\]
$ \implies {{s}_{2}}=\dfrac{10\times {{2}^{2}}}{2}=20m $
So, in 3 seconds distance covered by ball B is 5 m measured from the top, in both the cases the distances are measured from the top. To find out the distance between two balls we simply subtract them.
Now the distance between them after 3 seconds can be calculated as \[{{s}_{1}}-{{s}_{2}}\]
=(45-20)
=25m

So, the correct answer is “Option A”.

Note:
The body is moving downwards with the acceleration due to gravity and so the value of g is positive. If the case was just the opposite, that is the bodies were moving upwards then we would have taken the value of g to be negative.