At what angle of elevation , should a projectile be projected with velocity with velocity ` 20 ms^(-1)` , so as to reach a maximum height of 10 m ?

Options

A. `0^(@)`
B. ` 90^(@)`
C. `45^(@)`
D. `60^(@)`

Answer

To solve the problem of determining the angle of elevation at which a projectile should be launched to reach a maximum height of 10 m with an initial velocity of 20 m/s, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial velocity (u) = 20 m/s
- Maximum height (h) = 10 m
- Acceleration due to gravity (g) = 9.81 m/s² (approximately 10 m/s² for simplification)

2. **Use the formula for maximum height of a projectile:**
The maximum height (h) reached by a projectile is given by the formula:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]

3. **Substitute the known values into the formula:**
\[ 10 = \frac{(20)^2 \sin^2 \theta}{2 \times 10} \]
Simplifying this:
\[ 10 = \frac{400 \sin^2 \theta}{20} \]
\[ 10 = 20 \sin^2 \theta \]

4. **Rearranging the equation to solve for \(\sin^2 \theta\):**
\[ \sin^2 \theta = \frac{10}{20} = \frac{1}{2} \]

5. **Taking the square root to find \(\sin \theta\):**
\[ \sin \theta = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

6. **Finding the angle \(\theta\):**
The angle whose sine is \(\frac{\sqrt{2}}{2}\) is:
\[ \theta = 45^\circ \]

### Final Answer:
The angle of elevation at which the projectile should be projected is \(45^\circ\).

---