Question

A body is thrown vertically upwards from point \(A\), the top of a tower, and reaches the ground in time \(t_1\). If it is thrown vertically downwards from \(A\) with the same speed, it reaches the ground in time \(t_2\). If it is allowed to fall freely from \(A\), the time it takes to reach the ground is given by:

• A. \(t=\dfrac{t_1+t_2}{2}\)
• B. \(t=\dfrac{t_1-t_2}{2}\)
• C. \(t=\sqrt{t_1t_2}\)
• D. \(t=\sqrt{\dfrac{t_1}{t_2}}\)

Hint:

For motion under gravity from the same height with equal speeds: \[ t_{\text{free fall}}=\sqrt{t_{\uparrow}t_{\downarrow}} \] This result is independent of initial speed.

Answer 1

The Correct Option is C

Step 1: Let the height of the tower be \(h\) and acceleration due to gravity be \(g\).
Step 2: For upward throw: \[ h = ut_1 - \frac{1}{2}gt_1^2 \quad \cdots (1) \]
Step 3: For downward throw: \[ h = ut_2 + \frac{1}{2}gt_2^2 \quad \cdots (2) \]
Step 4: Adding (1) and (2): \[ 2h = u(t_1+t_2) - \frac{1}{2}g(t_1^2-t_2^2) \]
Step 5: Eliminating \(u\) and using free fall equation \(h=\frac{1}{2}gt^2\), we get: \[ t^2 = t_1t_2 \Rightarrow t=\sqrt{t_1t_2} \]