A pebbel is thrown horizantally from the top of a `20` m high tower with an initial velocity of `10 m//s`.The air drag is negligible . The speed of the peble when it is at the same distance from top as well as base of the tower `(g=10 m//s^2)`

Options

A. `10sqrt2 m//s`
B. `10sqrt3 m//s`
C. `20 m//s`
D. `25 m//s`

Explanation

To solve the problem, we need to find the speed of the pebble when it has fallen 10 meters vertically from the top of the tower. The pebble is thrown horizontally with an initial speed of 10 m/s, and we will consider both the horizontal and vertical components of its velocity.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- The height of the tower (h) = 20 m.
- The initial horizontal velocity (u_x) = 10 m/s.
- The initial vertical velocity (u_y) = 0 m/s (since it is thrown horizontally).
- The acceleration due to gravity (g) = 10 m/s².

2. **Determine the Distance Fallen:**
- The pebble falls 10 m vertically when it is at the same distance from the top and the base of the tower.

3. **Calculate the Final Vertical Velocity:**
- We can use the third equation of motion to find the final vertical velocity (v_y) after falling 10 m:
\[ v_y^2 = u_y^2 + 2gh \]
- Substituting the values:
\[ v_y^2 = 0 + 2 \cdot 10 \cdot 10 \]
\[ v_y^2 = 200 \]
\[ v_y = \sqrt{200} = 10\sqrt{2} \text{ m/s} \]

4. **Calculate the Resultant Velocity:**
- The resultant velocity (v) of the pebble when it has fallen 10 m can be found using the Pythagorean theorem, considering both horizontal and vertical components:
\[ v = \sqrt{u_x^2 + v_y^2} \]
- Substituting the values:
\[ v = \sqrt{(10)^2 + (10\sqrt{2})^2} \]
\[ v = \sqrt{100 + 200} = \sqrt{300} \]
\[ v = 10\sqrt{3} \text{ m/s} \]

5. **Final Answer:**
- The speed of the pebble when it is at the same distance from the top as well as the base of the tower is \( 10\sqrt{3} \) m/s.