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 a projectile fired at an angle, we can use the formula:

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

Where:
- \( R \) is the horizontal range,
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 1: Identify the given values
- Initial speed \( u = 20 \, \text{m/s} \)
- Angle of projection \( \theta = 60^\circ \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate \( \sin(2\theta) \)
Using the angle \( \theta = 60^\circ \):
\[ 2\theta = 120^\circ \]
Now, we need to find \( \sin(120^\circ) \):
\[ \sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

### Step 3: Substitute the values into the range formula
Now substituting the values into the range formula:
\[ R = \frac{(20 \, \text{m/s})^2 \cdot \sin(120^\circ)}{10 \, \text{m/s}^2} \]
\[ R = \frac{400 \cdot \frac{\sqrt{3}}{2}}{10} \]

### Step 4: 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:
\[ R = 20 \sqrt{3} \, \text{m} \]