An object is released from some height. Exactly after one second, another object is released from the same height. The distance between the two objects exactly after two seconds of the release of the second object will be:
A) 4.9m
B) 9.8m
C) 19.6m
D) 24.5m

Hint: Here the initial height of both the objects will remain the same. The weight of the ball is irrelevant as both the balls will go through the same acceleration (neglect air resistance and any other external environmental factors). Apply the equations of kinematics and solve.

Complete step by step solution:
Find the distance between the two balls:
Here we need to imagine two balls falling from a tower with the same height. One ball is falling 1 second earlier than the other ball and we have been asked to calculate the distance between the two objects at the end of second and after two seconds that means in total after 3 seconds, we have been asked to provide the distance between the two balls.
Apply the formula for kinematics:
$S = ut + \dfrac{1}{2}a{t^2}$;
Now, initially the ball is at rest when the ball is about to be dropped, so the initial velocity of the ball is zero.
$S = \dfrac{1}{2}a{t^2}$;
Now, the acceleration will be due to gravity so:
$ \Rightarrow S = \dfrac{1}{2}g{t^2}$;
Here, we are taking the time to be 3 seconds for the first ball. Put in the value in the above equation:
${S_1} = \dfrac{1}{2} \times 9.8 \times {3^2}$;
$ \Rightarrow {S_1} = 44.1m$;
Now, the second ball is one second late, so the time would be two seconds:
${S_2} = \dfrac{1}{2} \times 9.8 \times {2^2}$;
$ \Rightarrow {S_2} = 4.9 \times {2^2}$;
Do the needed calculation:
$ \Rightarrow {S_2} = 19.6m$;
Now, to find out the distance between the two balls, we know the distance that each ball covered from the initial point of drop, so if we subtract the distance covered by the first ball from the second ball, we would get the distance between the two balls.
$\Rightarrow {S_f} = {S_1} - {S_2}$;
$\Rightarrow {S_f} = 44.1 - 19.6$;
The distance between the two balls is:
$\Rightarrow {S_f} = 24.5m$;

Option (D) is correct. Therefore, the distance between the two objects exactly after two seconds of the release of the second object will be 24.5m.

Note: Here, read the question carefully as it is a conceptual question. You need to imagine that two balls are dropping from the same height with a time difference of 1second. The total time taken by the first ball is 1second + 2seconds and the second ball took only 2seconds. The rest is simple: apply the equation of kinematics and solve.