Four projectiles are projected with the same speed at angles `20^@,35^@,60^@ and 75^@`
with the horizontal. The range will be the maximum for the projectile whose angle of projection is

Options

A. `20^@`
B. `35^@`
C. `60^@`
D. `75^@`

Explanation

To determine which projectile has the maximum range when projected at angles of \(20^\circ\), \(35^\circ\), \(60^\circ\), and \(75^\circ\) with the same speed, we can use the formula for the range of a projectile:

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

where:
- \(R\) is the range,
- \(V\) is the initial speed,
- \(g\) is the acceleration due to gravity,
- \(\theta\) is the angle of projection.

### Step-by-Step Solution:

1. **Calculate the range for each angle**:
- For \(20^\circ\):
\[ R_1 = \frac{V^2 \sin(2 \times 20^\circ)}{g} = \frac{V^2 \sin(40^\circ)}{g} \]
- For \(35^\circ\):
\[ R_2 = \frac{V^2 \sin(2 \times 35^\circ)}{g} = \frac{V^2 \sin(70^\circ)}{g} \]
- For \(60^\circ\):
\[ R_3 = \frac{V^2 \sin(2 \times 60^\circ)}{g} = \frac{V^2 \sin(120^\circ)}{g} \]
- For \(75^\circ\):
\[ R_4 = \frac{V^2 \sin(2 \times 75^\circ)}{g} = \frac{V^2 \sin(150^\circ)}{g} \]

2. **Evaluate the sine values**:
- \(\sin(40^\circ)\)
- \(\sin(70^\circ)\)
- \(\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ)\)
- \(\sin(150^\circ) = \sin(180^\circ - 30^\circ) = \sin(30^\circ)\)

3. **Compare the sine values**:
- \(\sin(40^\circ) \approx 0.6428\)
- \(\sin(70^\circ) \approx 0.9397\)
- \(\sin(60^\circ) \approx 0.8660\)
- \(\sin(30^\circ) = 0.5\)

4. **Determine the maximum range**:
- From the calculated sine values, we can see that:
\[ \sin(70^\circ) > \sin(60^\circ) > \sin(40^\circ) > \sin(30^\circ) \]
- Therefore, the maximum range will be for the angle \(35^\circ\).

### Conclusion:
The projectile with the angle of projection \(35^\circ\) will have the maximum range.

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