Question

From the top of a tower of height $200\text{m}$, a ball A is projected up with $10{\text{ms}}^{-1}$ and $2\text{s}$ later another ball B is projected vertically down with the same speed. Then (Neglect air friction)

• A. Both A and B will reach the ground simultaneously
• B. Ball A will hit the ground, $2\text{s}$ after B hits the ground
• C. Both the balls will hit the ground with the same velocity
• D. Both the balls will hit the ground with the different velocity

Answer 1

Answer: (A), (C)

Both the balls will hit the ground with the same velocity

Ball A will return to the top of tower after
$T=\frac{2u}{g}=\frac{2\times 10}{10}=2\text{s}$
with a speed of $10\text{m}{s}^{-1}$ downward.
And at the same moment B is also projected downward with $10\text{m}{s}^{-1}.$
So both reach ground simultaneously. Also they will hit the ground with same speed.