A child stands on the edge of the cliff `10m` above the ground and throws a stone horizontally with an initial speed of `5ms^(-1)`.Neglecting the air resistance,the speed with which the stone hits the ground will be_________`ms^(-1)`.

Options

A. `15`
B. `20`
C. `30`
D. `25`

Answer

To solve the problem, we need to determine the speed with which the stone hits the ground after being thrown horizontally from a height of 10 meters. We will consider both the horizontal and vertical components of the stone's motion.

### Step-by-Step Solution:

1. **Identify the Components of Motion:**
- The stone is thrown horizontally with an initial speed \( v_x = 5 \, \text{m/s} \).
- The vertical motion is influenced by gravity, with an initial vertical speed \( v_y = 0 \, \text{m/s} \).

2. **Calculate the Time of Flight:**
- The time \( t \) it takes for the stone to fall from a height \( h = 10 \, \text{m} \) can be calculated using the formula for free fall:
\[ h = \frac{1}{2} g t^2 \]
- Rearranging gives:
\[ t = \sqrt{\frac{2h}{g}} \]
- Using \( g = 10 \, \text{m/s}^2 \):
\[ t = \sqrt{\frac{2 \times 10}{10}} = \sqrt{2} \, \text{s} \]

3. **Calculate the Vertical Velocity at Impact:**
- The vertical velocity \( v_y \) just before impact can be calculated using:
\[ v_y = g t \]
- Substituting the values:
\[ v_y = 10 \times \sqrt{2} \approx 14.14 \, \text{m/s} \]

4. **Calculate the Resultant Velocity:**
- The resultant velocity \( v \) when the stone hits the ground can be found using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} \]
- Substituting the values:
\[ v = \sqrt{(5)^2 + (10\sqrt{2})^2} \]
\[ v = \sqrt{25 + 200} = \sqrt{225} = 15 \, \text{m/s} \]

5. **Final Answer:**
- The speed with which the stone hits the ground is \( 15 \, \text{m/s} \).

### Summary:
The speed with which the stone hits the ground is **15 m/s**.