In the question number 62. the maximum height attained by the ball is

Options

A. 11.25 m
B. 48.2 m
C. 23. 5 m
D. 68 m

Explanation

To solve the problem of finding the maximum height attained by a ball thrown at an angle, we can use the formula for maximum height in projectile motion:

### Step-by-Step Solution:

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

2. **Use the Formula for Maximum Height:**
The formula for maximum height \( H \) attained by a projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

3. **Calculate \( \sin \theta \):**
For \( \theta = 30^\circ \):
\[ \sin 30^\circ = \frac{1}{2} \]

4. **Substitute the Values into the Formula:**
First, calculate \( \sin^2 \theta \):
\[ \sin^2 30^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
Now substitute \( u \), \( \sin^2 \theta \), and \( g \) into the height formula:
\[ H = \frac{(30)^2 \cdot \frac{1}{4}}{2 \cdot 10} \]

5. **Calculate \( H \):**
\[ H = \frac{900 \cdot \frac{1}{4}}{20} = \frac{225}{20} = 11.25 \, \text{m} \]

6. **Conclusion:**
The maximum height attained by the ball is \( 11.25 \, \text{m} \).