Question

Five stones are dropped successively from a height of 50 m from the ground with a time interval of half a second between two successive stones. The relative velocity between the first and third stones when they are in motion is \[ (g=10~\text{m s}^{-2}) \]

• A. \(15~\text{m s}^{-1}\)
• B. \(5~\text{m s}^{-1}\)
• C. \(10~\text{m s}^{-1}\)
• D. \(20~\text{m s}^{-1}\)

Hint:

For bodies dropped at different times, \[ \boxed{ v_{\text{rel}}=g(\Delta t) } \] where \(\Delta t\) is the difference in release times.

Answer 1

The Correct Option is C

Step 1: Determine the time difference. The third stone is released \[ 2\times 0.5=1~\text{s} \] after the first stone.

Step 2: Write the velocities. If the first stone has been falling for \(t\) seconds, \[ v_1=gt \] The third stone has been falling for \[ (t-1)\ \text{s}, \] so \[ v_3=g(t-1). \]

Step 3: Find the relative velocity. \[ v_{\text{rel}} =v_1-v_3 =gt-g(t-1) =g =10~\text{m s}^{-1}. \] Hence, \[ \boxed{v_{\text{rel}}=10~\text{m s}^{-1}} \] Therefore, \[ \boxed{(C)} \] is the correct answer.