Question

Two stones begin to fall from rest from the same height, with the second stone starting to fall \( t_1 \) seconds after the first falls from rest. The distance of separation between the two stones becomes \( H \) after the first stone starts its motion. Then \( t_1 \) is equal to:

• A. \( \frac{H}{2g} \)
• B. \( \frac{H}{g} \)
• C. \( \frac{H}{3g} \)
• D. \( \frac{H}{4g} \)

Hint:

When dealing with problems involving free fall, always keep track of the times and distances involved. Use the equations of motion to find the relationship between them.

Answer 1

The Correct Option is A

Let's consider the motion of the two stones. The first stone falls under gravity, and its distance fallen after time \( t \) is given by: \[ s_1 = \frac{1}{2} g t^2 \] The second stone falls \( t_1 \) seconds later, and the distance it falls is: \[ s_2 = \frac{1}{2} g (t_1)^2 \] The distance between the two stones is: \[ \Delta s = s_1 - s_2 = \frac{1}{2} g (t^2 - t_1^2) \] Given that the distance of separation between the two stones is \( H \), we have: \[ H = \frac{1}{2} g (t^2 - t_1^2) \] For the separation to be equal to \( H \), the time \( t_1 \) is: \[ t_1 = \sqrt{\frac{H}{2g}} \] Thus, \( t_1 = \frac{H}{2g} \).