From the ground, a projectile is fired at an angle of 60 degree to the horizontal with a speed of `20" m s"^(-1)` The horizontal range of the projectile is

Options

A. `10sqrt3`
B. 20 m
C. `20sqrt3m`
D. `40sqrt3`

Answer

To find the horizontal range of the projectile, we can use the formula for the range \( R \) of a projectile launched at an angle \( \theta \) with an initial speed \( u \):

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

Where:
- \( u \) is the initial speed,
- \( \theta \) is the launch angle,
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \) for this problem).

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Initial speed \( u = 20 \, \text{m/s} \)
- Launch angle \( \theta = 60^\circ \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

2. **Calculate \( 2\theta \)**:
\[ 2\theta = 2 \times 60^\circ = 120^\circ \]

3. **Calculate \( \sin(2\theta) \)**:
- We need to find \( \sin(120^\circ) \).
- \( \sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \)

4. **Substitute Values into the Range Formula**:
\[ R = \frac{(20 \, \text{m/s})^2 \cdot \sin(120^\circ)}{10 \, \text{m/s}^2} \]
\[ R = \frac{400 \cdot \frac{\sqrt{3}}{2}}{10} \]

5. **Simplify the Expression**:
\[ R = \frac{400 \cdot \frac{\sqrt{3}}{2}}{10} = \frac{400\sqrt{3}}{20} = 20\sqrt{3} \, \text{m} \]

### Final Answer:
The horizontal range of the projectile is \( R = 20\sqrt{3} \, \text{meters} \).

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