Q. From a building two balls $A$ to $B$ are thrown such that $A$ is thrown upwards and $B$ downwards (both vertically). If $v_{A}$ and $v_{B}$ are their respective velocities on reaching the ground, then
A) $v_{B} > v_{A}$
B) $v_{B} = v_{A}$
C) $v_{A} > v_{B}$
D) Their velocities depends on their masses
Explanation
(b) From conservation of energy,
Potential energy at height $h =$ kinetic energy at ground
Therefore, at height $h$, potential energy of ball $A$
PE = $m_{A}gh$
KE at ground $= \frac{1}{2}m_{A}v_{A}^{2}$
So, $m_{A}gh = \frac{1}{2}m_{A}v_{A}^{2}$
$v_{A} = \sqrt{2gh}$
Similarly, $v_{B} = \sqrt{2gh}$
Therefore, $v_{A} = v_{B}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.