Question

A man standing on the edge of the terrace of a high rise building throws a stone, vertically up with a speed of 20 m/s. Two seconds later, an identical stone is thrown vertically downwards with the same speed of 20 m/s. Then :

• A. the relative velocity between the two stones remains 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 s
• D. if the collision on the ground is perfectly elastic, both will rise to the same height above the ground

Answer 1

Answer: (A), (B), (D)

The correct options are
A both will have the same kinetic energy, when they hit the ground
B the relative velocity between the two stones remains constant till one hits the ground
D if the collision on the ground is perfectly elastic, both will rise to the same height above the ground

Since when the two stones are in air, they have the same acceleration(g), thus their relative speed remains same. Third equation of motion says:
$v^2-u^2=2as$
for both the stones initial speed(u = 20 m/s), displacement(height of the building) and acceleration(g) are same respectively. Thus their final speed is also same. Since the particles are identical(masses are same), they will have the same kinetic energy.
for first stone time taken to reach the ground is given by the equation:
$H=-20t_1+\frac{g{t}_1^2}{2}$ .......(1)
for second stone time taken to reach the ground is given by the equation:
$H=20t_2+\frac{g{t}_2^2}{2}$ .......(2)
(1)-(2) gives
$-20(t_1+t_2)+\frac{g{t}_1^2-g{t}_2^2}{2}=0\Rightarrow (t_1+t_2)(-20+\frac{g(t_1-t_2)}{2})=0\Rightarrow (t_1-t_2)=\frac{40}{g}\ne 2$
since the option A is correct, the velocity with which they reach the ground is same and the collision is elastic, their velocities just after the collision will be same. Thus, they reach to the same height.