A bullet is fired in a horizontal direction from a tower while a stone is simultaneously dropped from the same point then :

Options

A. The bullet and the stone will reach the ground simultaneously
B. The stone will reach earlier
C. The bullet will reach earlier
D. None of these

Answer

To solve the problem, we need to analyze the motion of both the bullet and the stone.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A bullet is fired horizontally from a tower, and a stone is dropped from the same height at the same time. We need to determine which one reaches the ground first.

2. **Identify the Variables**:
- Let the height of the tower be \( H \).
- The initial vertical velocity of the stone is \( U_{\text{stone}} = 0 \) (since it is dropped).
- The initial vertical velocity of the bullet is also \( U_{\text{bullet}} = 0 \) in the vertical direction (it only has horizontal velocity).

3. **Vertical Motion Analysis**:
- For both the bullet and the stone, the only force acting on them in the vertical direction is gravity. Therefore, they both experience the same acceleration \( g \) downwards.
- The equation of motion for vertical displacement is given by:
\[ H = U_{\text{initial}} t + \frac{1}{2} g t^2 \]
- For the stone:
\[ H = 0 \cdot t + \frac{1}{2} g t^2 \implies H = \frac{1}{2} g t^2 \]
- For the bullet, since it is fired horizontally, its vertical motion is also governed by the same equation:
\[ H = 0 \cdot t + \frac{1}{2} g t^2 \implies H = \frac{1}{2} g t^2 \]

4. **Time to Reach the Ground**:
- From the equation \( H = \frac{1}{2} g t^2 \), we can solve for \( t \):
\[ t = \sqrt{\frac{2H}{g}} \]
- This time \( t \) is the same for both the stone and the bullet since they both fall from the same height \( H \) and are subject to the same gravitational acceleration \( g \).

5. **Conclusion**:
- Since both the bullet and the stone take the same time \( t \) to reach the ground, they will hit the ground simultaneously.

### Answer:
The bullet and the stone will reach the ground simultaneously.