Q. From the top of tower, a stone is thrown up. It reaches the ground in $t_{1}$ second. A second stone thrown down with the same speed reaches the ground in $t_{2}$ second. A third stone released from rest reaches the ground in $t_{3}$ second. Then
A) $t_{3} = \frac{\left( t_{1} + t_{2} \right)}{2}$
B) $t_{3} = \sqrt{t_{1}t_{2}}$
C) $\frac{1}{t_{3}} = \frac{1}{t_{1}} - \frac{1}{t_{2}}$
D) $t_{3}^{2} = t_{2}^{2} - t_{1}^{2}$
Explanation
The correct answer is Option B: $t_{3} = \sqrt{t_{1}t_{2}}$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.