In the question number 62, the distance from the thrower to the point where the ball returns to the same level is

Options

A. 58 m
B. 68 m
C. 78 m
D. 88 m

Explanation

To solve the problem of finding the distance from the thrower to the point where the ball returns to the same level, we will use the formula for the range of a projectile. Here are the steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial velocity (u) = 30 m/s
- Angle of projection (θ) = 30 degrees
- Acceleration due to gravity (g) = 10 m/s²

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

3. **Calculate \( \sin(2\theta) \):**
- First, calculate \( 2\theta \):
\[ 2\theta = 2 \times 30^\circ = 60^\circ \]
- Now, find \( \sin(60^\circ) \):
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

4. **Substitute the values into the range formula:**
\[ R = \frac{(30 \, \text{m/s})^2 \cdot \sin(60^\circ)}{10 \, \text{m/s}^2} \]
\[ R = \frac{900 \cdot \frac{\sqrt{3}}{2}}{10} \]

5. **Simplify the expression:**
\[ R = \frac{900 \cdot \frac{\sqrt{3}}{2}}{10} = \frac{900 \sqrt{3}}{20} = 45 \sqrt{3} \]

6. **Calculate the numerical value:**
- Using \( \sqrt{3} \approx 1.732 \):
\[ R \approx 45 \times 1.732 \approx 77.94 \, \text{m} \]

7. **Round to the nearest whole number:**
\[ R \approx 78 \, \text{m} \]

### Final Answer:
The distance from the thrower to the point where the ball returns to the same level is approximately **78 meters**.