From a 10 m high building a ston 'A' is dropped, and simultaneously another identical stone 'B' is thrown horizontally with an initial speed of `5ms^(-1)`. Which one of the following statement is true?

Options

A. It is not possible to calculate which one of the two stones will reach the ground first
B. Both the stones ('A' and 'B') will reach the ground simultaneously
C. A' stone reaches the ground earlier than 'B'
D. B' stone reaches the ground earlier than 'A'

Answer

To solve the problem, we need to analyze the motion of both stones A and B.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Height of the building (h) = 10 m
- Initial speed of stone B (horizontal) = 5 m/s
- Initial speed of stone A (dropped) = 0 m/s

2. **Determine Time of Flight for Stone A**:
- Since stone A is dropped, its initial vertical velocity (u_y) = 0 m/s.
- The equation of motion for vertical displacement is given by:
\[ s = u_y t + \frac{1}{2} a t^2 \]
- Here, \(s = -10\) m (downward), \(u_y = 0\), and \(a = -g\) (acceleration due to gravity, approximately 9.8 m/s²).
- Plugging in the values:
\[ -10 = 0 \cdot t + \frac{1}{2} (-g) t^2 \]
\[ -10 = -\frac{1}{2} g t^2 \]
\[ 10 = \frac{1}{2} g t^2 \]
\[ t^2 = \frac{20}{g} \]
\[ t = \sqrt{\frac{20}{g}} \]

3. **Determine Time of Flight for Stone B**:
- Stone B is thrown horizontally, so its initial vertical velocity (u_y) = 0 m/s.
- The vertical motion is the same as stone A since both stones fall from the same height:
- Using the same equation of motion:
\[ -10 = 0 \cdot t + \frac{1}{2} (-g) t^2 \]
- This gives the same result as for stone A:
\[ t = \sqrt{\frac{20}{g}} \]

4. **Conclusion**:
- Both stones A and B take the same time to reach the ground, which is \(t = \sqrt{\frac{20}{g}}\).
- Therefore, they will hit the ground simultaneously.

### Answer:
Both stones A and B will reach the ground simultaneously.