A ball is thrown at a speed of 20 m/s at an angle of `30 ^@` with the horizontal . The maximum height reached by the ball is
(Use `g=10 m//s^2`)

Options

A. 2 m
B. 3 m
C. 4 m
D. 5 m

Answer

To find the maximum height reached by a ball thrown at a speed of 20 m/s at an angle of 30 degrees with the horizontal, we can use the formula for maximum height in projectile motion:

\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

Where:
- \( H \) = maximum height
- \( u \) = initial speed (20 m/s)
- \( \theta \) = angle of projection (30 degrees)
- \( g \) = acceleration due to gravity (10 m/s²)

### Step-by-Step Solution:

1. **Identify the values:**
- Initial speed, \( u = 20 \, \text{m/s} \)
- Angle of projection, \( \theta = 30^\circ \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate \( \sin \theta \):**
- For \( \theta = 30^\circ \), \( \sin 30^\circ = \frac{1}{2} \)
- Therefore, \( \sin^2 30^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)

3. **Substitute the values into the formula:**
\[ H = \frac{(20)^2 \cdot \sin^2 30^\circ}{2 \cdot 10} \]
\[ H = \frac{400 \cdot \frac{1}{4}}{20} \]

4. **Simplify the expression:**
\[ H = \frac{400 \cdot \frac{1}{4}}{20} = \frac{100}{20} = 5 \, \text{m} \]

5. **Conclusion:**
- The maximum height reached by the ball is \( H = 5 \, \text{m} \).

### Final Answer:
The maximum height reached by the ball is **5 meters**.