One stone is projected horizontally from a `20 m` high cliff with an initial speed of `10 ms^(-1)`. A second stone is simultaneously dropped from that cliff. Which of the following is true?

Options

A. Both strike the ground with same speed.
B. Tha ball with initial speed `10ms^(-1)` reaches the ground first.
C. The ball which is dropped, reaches the ground first.
D. Both strike the ground with different speed

Explanation

To solve the problem, we need to analyze the motion of both stones: one that is projected horizontally and the other that is dropped vertically.

### Step-by-Step Solution:

1. **Identify the given data:**
- Height of the cliff (h) = 20 m
- Initial speed of the horizontally projected stone (u₁) = 10 m/s
- Initial speed of the dropped stone (u₂) = 0 m/s (since it is dropped)

2. **Determine the time taken for both stones to reach the ground:**
- For the stone that is dropped (second stone), we can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( s = -h = -20 \, \text{m} \), \( u = 0 \), and \( a = g \approx 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).
\[ -20 = 0 \cdot t_2 + \frac{1}{2} (-g) t_2^2 \]
This simplifies to:
\[ -20 = -\frac{1}{2} g t_2^2 \]
Rearranging gives:
\[ t_2^2 = \frac{40}{g} \]
Thus:
\[ t_2 = \sqrt{\frac{40}{g}} \]

3. **For the stone that is projected horizontally (first stone):**
- The vertical motion is the same as that of the dropped stone because both stones fall the same vertical distance under the influence of gravity. Therefore, we can use the same equation:
\[ -20 = 0 \cdot t_1 + \frac{1}{2} (-g) t_1^2 \]
This gives:
\[ t_1^2 = \frac{40}{g} \]
Thus:
\[ t_1 = \sqrt{\frac{40}{g}} \]

4. **Compare the times:**
- Since both \( t_1 \) and \( t_2 \) are equal:
\[ t_1 = t_2 \]
- Therefore, both stones hit the ground at the same time.

5. **Determine the speeds at which they hit the ground:**
- The speed of the dropped stone just before it hits the ground can be calculated using:
\[ v = u + gt \]
For the dropped stone:
\[ v_2 = 0 + g t_2 = g \sqrt{\frac{40}{g}} = \sqrt{40g} \]
- The speed of the horizontally projected stone just before it hits the ground can be calculated using:
\[ v = u + gt \]
For the horizontally projected stone:
\[ v_1 = 10 + g t_1 = 10 + g \sqrt{\frac{40}{g}} = 10 + \sqrt{40g} \]

6. **Conclusion:**
- Both stones strike the ground at the same time, but with different speeds.

### Final Answer:
Both stones strike the ground at the same time, but they do not have the same speed.