A ball is thrown upward at an angle of `45^(@)` with the horizontal and lands on the top edge of a builing that is 20 m away. The top edge is 10 m above the throwing point. The initial speed of the ball in metre/second is (take `g =10 m//s^(2)`)

Options

A. 10 m/s
B. 20 m/s
C. 25 m/s
D. 30 m/s

Explanation

To solve the problem, we need to find the initial speed of a ball thrown at an angle of 45 degrees that lands on the top edge of a building 20 meters away and 10 meters high. We will use the equations of projectile motion to derive the solution.

### Step-by-Step Solution:

1. **Understand the Problem**:
- The ball is thrown at an angle of \( 45^\circ \).
- The horizontal distance to the building (range, \( x \)) is 20 m.
- The vertical height of the building (height, \( y \)) is 10 m.
- We need to find the initial speed \( u \).

2. **Use the Equation of Trajectory**:
The equation of trajectory for projectile motion is given by:
\[ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \]
where:
- \( y \) is the vertical displacement (10 m),
- \( x \) is the horizontal displacement (20 m),
- \( \theta \) is the angle of projection (45 degrees),
- \( g \) is the acceleration due to gravity (10 m/s²),
- \( u \) is the initial speed.

3. **Substituting Values**:
For \( \theta = 45^\circ \):
- \( \tan 45^\circ = 1 \)
- \( \cos 45^\circ = \frac{1}{\sqrt{2}} \)

Substitute these values into the equation:
\[ 10 = 20 \cdot 1 - \frac{10 \cdot 20^2}{2 u^2 \left(\frac{1}{\sqrt{2}}\right)^2} \]

4. **Simplifying the Equation**:
The equation simplifies to:
\[ 10 = 20 - \frac{10 \cdot 400}{2 u^2 \cdot \frac{1}{2}} \]
\[ 10 = 20 - \frac{4000}{u^2} \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{4000}{u^2} = 20 - 10 \]
\[ \frac{4000}{u^2} = 10 \]

6. **Solving for \( u^2 \)**:
Multiply both sides by \( u^2 \):
\[ 4000 = 10 u^2 \]
\[ u^2 = \frac{4000}{10} = 400 \]

7. **Finding \( u \)**:
Taking the square root of both sides:
\[ u = \sqrt{400} = 20 \text{ m/s} \]

### Final Answer:
The initial speed of the ball is \( 20 \text{ m/s} \).