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))`

Answer

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

### Step-by-Step Solution:

1. **Identify the Variables:**
- Horizontal distance to the building, \( x = 20 \, \text{m} \)
- Height of the building, \( y = 10 \, \text{m} \)
- Angle of projection, \( \theta = 45^\circ \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Use the Projectile Motion Equation:**
The equation for the height \( y \) of a projectile at a horizontal distance \( x \) is given by:
\[ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \]
Here, \( u \) is the initial speed we need to find.

3. **Substitute the Known Values:**
- Since \( \theta = 45^\circ \), we know that \( \tan 45^\circ = 1 \) and \( \cos 45^\circ = \frac{1}{\sqrt{2}} \).
- Substitute \( y = 10 \), \( x = 20 \), \( g = 10 \), and \( \theta = 45^\circ \) into the equation:
\[ 10 = 20 \cdot 1 - \frac{10 \cdot 20^2}{2 u^2 \cdot \left(\frac{1}{\sqrt{2}}\right)^2} \]

4. **Simplify the Equation:**
- Calculate \( \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \).
- Substitute this into the equation:
\[ 10 = 20 - \frac{10 \cdot 400}{2 u^2 \cdot \frac{1}{2}} \]
- This simplifies to:
\[ 10 = 20 - \frac{4000}{u^2} \]

5. **Rearrange the Equation:**
- Move \( 10 \) to the right side:
\[ 10 = 20 - \frac{4000}{u^2} \implies \frac{4000}{u^2} = 10 \]

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

7. **Calculate \( u \):**
- Take the square root of both sides to find \( u \):
\[ u = \sqrt{400} = 20 \, \text{m/s} \]

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

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