A ball is projected at an angle 30° with the horizontal with the velocity 49`ms^(-1)`. The horizontal range is

Options

A. 122.5 m
B. 245 m
C. `245sqrt(3)` m
D. `122.5 sqrt(3)` m

Answer

To find the horizontal range of a ball projected at an angle, we can use the formula for horizontal range \( R \):

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

where:
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

### Step-by-step Solution:

1. **Identify the given values**:
- Initial velocity \( u = 49 \, \text{m/s} \)
- Angle of projection \( \theta = 30^\circ \)
- Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)

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

3. **Substitute values into the range formula**:
\[ R = \frac{(49)^2 \cdot \sin 60^\circ}{g} \]
\[ R = \frac{49^2 \cdot \frac{\sqrt{3}}{2}}{9.8} \]

4. **Calculate \( 49^2 \)**:
\[ 49^2 = 2401 \]

5. **Substitute \( 49^2 \) into the equation**:
\[ R = \frac{2401 \cdot \frac{\sqrt{3}}{2}}{9.8} \]

6. **Simplify the expression**:
\[ R = \frac{2401 \sqrt{3}}{2 \cdot 9.8} \]
\[ R = \frac{2401 \sqrt{3}}{19.6} \]

7. **Calculate \( \frac{2401}{19.6} \)**:
\[ \frac{2401}{19.6} = 122.5 \]

8. **Final result**:
\[ R = 122.5 \sqrt{3} \, \text{meters} \]

Thus, the horizontal range is \( 122.5 \sqrt{3} \, \text{meters} \).