A ball is thrown horizontally with a speed of `20 m//s` from the top of a tower of height 100m
velocity with which the ball strikes the ground is .

Options

A. `20 sqrt6 m//s` at angle `tan ^(-10 (sqrt7)` below the horizontal
B. `20 sqrt6 m//s` at an angle `tan ^(-1) (sqrt6)` below the horizontal
C. `20 sqrt6 m//s` at an angle `tan ^(-1)(sqrt5)` below the horizontal
D. `10 sqrt6 m//s` at an angle `tan ^(-1) (sqrt5)` below the horizontal

Explanation

To solve the problem of finding the velocity with which the ball strikes the ground when thrown horizontally from a height of 100 m, we can break the solution into several steps.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Initial horizontal velocity, \( u_x = 20 \, \text{m/s} \)
- Height of the tower, \( h = 100 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Time of Flight:**
The time \( t \) it takes for the ball to fall to the ground can be calculated using the formula:
\[ t = \sqrt{\frac{2h}{g}} \]
Substituting the values:
\[ t = \sqrt{\frac{2 \times 100}{10}} = \sqrt{20} = 2\sqrt{5} \, \text{s} \]

3. **Calculate the Final Velocity in the Y-Direction:**
The final velocity \( v_y \) in the vertical direction can be calculated using the equation:
\[ v_y = u_y + g \cdot t \]
Since the initial vertical velocity \( u_y = 0 \):
\[ v_y = 0 + 10 \cdot (2\sqrt{5}) = 20\sqrt{5} \, \text{m/s} \]

4. **Determine the Final Velocity in the X-Direction:**
The horizontal velocity remains constant since there is no horizontal acceleration:
\[ v_x = u_x = 20 \, \text{m/s} \]

5. **Calculate the Magnitude of the Resultant Velocity:**
The resultant velocity \( v \) when the ball strikes the ground can be found using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the values:
\[ v = \sqrt{(20)^2 + (20\sqrt{5})^2} = \sqrt{400 + 400 \cdot 5} = \sqrt{400 + 2000} = \sqrt{2400} \]
Simplifying:
\[ v = \sqrt{2400} = 20\sqrt{6} \, \text{m/s} \]

6. **Calculate the Angle of Impact:**
The angle \( \theta \) with respect to the horizontal can be calculated using:
\[ \tan \theta = \frac{v_y}{v_x} = \frac{20\sqrt{5}}{20} = \sqrt{5} \]
Thus,
\[ \theta = \tan^{-1}(\sqrt{5}) \]

### Final Answer:
The velocity with which the ball strikes the ground is \( 20\sqrt{6} \, \text{m/s} \) and the angle of impact is \( \tan^{-1}(\sqrt{5}) \) below the horizontal.