A body is projected at angle 30° to horizontal with a velocity 50 `ms^(-1)` .In the above problem range of body is m. [g = 10 `ms^(-2)` ]

Options

A. `125sqrt(3)`
B. `250 sqrt(3)`
C. 125
D. 250

Answer

To find the range of a projectile launched at an angle, we can use the formula for the range \( R \) of a projectile, which is given by:

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

where:
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

### Step-by-step Solution:

1. **Identify the given values:**
- Initial velocity, \( u = 50 \, \text{m/s} \)
- Angle of projection, \( \theta = 30^\circ \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

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 the values into the range formula:**
\[ R = \frac{u^2 \sin(2\theta)}{g} = \frac{(50)^2 \sin(60^\circ)}{10} \]

4. **Calculate \( u^2 \):**
\[ u^2 = 50^2 = 2500 \]

5. **Substitute \( u^2 \) and \( \sin(60^\circ) \) into the equation:**
\[ R = \frac{2500 \cdot \frac{\sqrt{3}}{2}}{10} \]

6. **Simplify the expression:**
- First, calculate \( \frac{2500}{10} = 250 \):
\[ R = 250 \cdot \frac{\sqrt{3}}{2} \]
- Now, calculate \( 250 \cdot \frac{1}{2} = 125 \):
\[ R = 125\sqrt{3} \]

7. **Final Result:**
- The range of the body is:
\[ R = 125\sqrt{3} \, \text{meters} \]

### Conclusion:
The range of the projectile is \( 125\sqrt{3} \, \text{meters} \).