Question

A man standing on the edge of the terrace of a high rise building throws a stone, vertically up with a speed of $20m/s$. Two second later, an identical stone is thrown vertically downwards with the same speed of $20m/s$. Then:
A. the relative velocity between the two stones remain constant till one hits the ground
B. both will have the same kinetic energy when they hit the ground
C. the time interval between their hitting the ground is 2 seconds
D. if the collisions on the ground are perfectly elastic both will rise to the same height above the ground.

Answer 1

Hint: Using the equation of motion, \[{v^2} - {u^2} = 2as\] and \[h = ut + \dfrac{1}{2}a{t^2}\]. Since a stone thrown upwards its velocity will decrease so we take minus sign on its initial velocity.

 Complete step-by-step solution:
When the two stones thrown in air at same time in air relative velocity of both stones will be the same.
Now from third equation of motion:
\[{v^2} - {u^2} = 2as\]
Given for both stones initial speed,$u = 20m/s$. Displacement of stones is the height of the building (\[h\]) and acceleration,\[g\]is the same for both. Thus the final velocity of stone will also be the same. Since stones are identical, they will have the same kinetic energy. Also let time taken by first stone be \[{t_1}\] and time taken by second stone be \[{t_2}\].
We know, \[h = ut + \dfrac{1}{2}a{t^2}\] putting the given values in this equation:
Time taken by the first stone to reach the ground is given by:
\[h = - 20{t_1} + \dfrac{1}{2}gt_1^2\] -------(1)
Time taken by the second stone to reach the ground is given by:
\[h = 20{t_2} + \dfrac{1}{2}gt_2^2\]--------(2)
On equating (1) and (2), we get: \[ - 20{t_1} + \dfrac{1}{2}gt_1^2 = 20{t_2} + \dfrac{1}{2}gt_2^2\]
On further solving, \[20{t_1} + 20{t_2} + \dfrac{1}{2}gt_2^2 - \dfrac{1}{2}gt_1^2 = 0\]
On solving, \[20({t_1} + {t_2}) - \dfrac{1}{2}g(t_1^2 - t_2^2) = 0\] \[ \Rightarrow ({t_1} + {t_2})[20 - \dfrac{1}{2}g({t_1} - {t_2})] = 0\]
On further solving, \[ \Rightarrow ({t_1} - {t_2}) = \dfrac{{40}}{g} \ne 2\]
Since option A is correct, the velocity with which they reach the ground is the same and the collision is elastic, their velocities just after the collision will be the same. Thus, they reach the same height.

Note:- Only option A,B and D are correct. Option C is not correct because the time interval is not equal to 2 sec. Take precautions during solving the mathematical portion since it is very complicated. Take care signs when transferring the values across the equation.