A ball is drpped from rest at height `4h`. After it has fallen a distance `d`, a second ball is dropped from rest at height `h`. What should `d` be (in terms of `h`) so that the balls hit the ground at the same time?

Options

A. `d=(3H)/2`
B. `d=H/3`
C. `d=H`
D. `d=3H`

Explanation

To solve the problem, we need to find the distance \( d \) that the first ball falls before the second ball is dropped, such that both balls hit the ground at the same time.

### Step-by-step Solution:

1. **Identify the heights and times**:
- The first ball is dropped from a height of \( 4h \).
- The second ball is dropped from a height of \( h \).
- Let \( T \) be the time taken by the first ball to hit the ground, and \( t \) be the time taken by the second ball to hit the ground after it is dropped.

2. **Use the equation of motion for the first ball**:
- The first ball falls a distance of \( 4h \) under gravity. The equation of motion is:
\[ s = ut + \frac{1}{2} g T^2 \]
Since the ball is dropped from rest, \( u = 0 \):
\[ 4h = \frac{1}{2} g T^2 \]
Rearranging gives:
\[ T^2 = \frac{8h}{g} \]

3. **Use the equation of motion for the second ball**:
- The second ball falls a distance of \( h \) under gravity. The time it falls is \( t \):
\[ h = \frac{1}{2} g t^2 \]
Rearranging gives:
\[ t^2 = \frac{2h}{g} \]

4. **Relate the times**:
- The first ball falls for a total time \( T \), while the second ball falls for a time \( t \). Since the second ball is dropped after the first ball has fallen a distance \( d \), we have:
\[ T = t + \text{time taken to fall } d \]
- The time taken for the first ball to fall a distance \( d \) is:
\[ d = \frac{1}{2} g (T - t)^2 \]
- We know \( T = 2t \) from the previous steps (since \( T^2 = 4t^2 \)). Thus:
\[ d = \frac{1}{2} g (T - t)^2 = \frac{1}{2} g (2t - t)^2 = \frac{1}{2} g t^2 \]

5. **Substitute \( t^2 \)**:
- From the earlier equation for the second ball, we have \( t^2 = \frac{2h}{g} \):
\[ d = \frac{1}{2} g \left(\frac{2h}{g}\right) = h \]

### Final Result:
Thus, the distance \( d \) that the first ball must fall before the second ball is dropped, so that both hit the ground at the same time, is:
\[ d = h \]