A body is projected at an angle of `30^(@)` with a velocity of `320 ft//"sec"`, then
the range on horizontal plane ...........

Answer

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

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

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 \( u = 320 \, \text{ft/sec} \)
- Angle of projection \( \theta = 30^\circ \)
- Acceleration due to gravity \( g = 32 \, \text{ft/sec}^2 \)

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

3. **Substitute the values into the range formula:**
\[ R = \frac{(320)^2 \cdot \sin(60^\circ)}{32} \]

4. **Calculate \( (320)^2 \):**
\[ (320)^2 = 102400 \]

5. **Substitute \( \sin(60^\circ) \) into the equation:**
\[ R = \frac{102400 \cdot \frac{\sqrt{3}}{2}}{32} \]

6. **Simplify the expression:**
- First, simplify \( \frac{102400}{32} \):
\[ \frac{102400}{32} = 3200 \]
- Now substitute back:
\[ R = 3200 \cdot \frac{\sqrt{3}}{2} = 1600\sqrt{3} \]

7. **Final Result:**
\[ R \approx 1600 \cdot 1.732 \approx 2764.8 \, \text{ft} \]

### Conclusion:
The range on the horizontal plane is \( 1600\sqrt{3} \, \text{ft} \) or approximately \( 2764.8 \, \text{ft} \).