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. (a)`20^@`
B. (b)`35^@`
C. (c)`60^@`
D. (d)`75^@`

Explanation

To determine which projectile has the maximum range, we can use the formula for the range of a projectile, which is given by:

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

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

Since all projectiles are projected with the same speed, we can focus on maximizing \( \sin(2\theta) \).

### Step-by-Step Solution:

1. **Calculate \( \sin(2\theta) \) for each angle:**
- For \( \theta = 20^\circ \):
\[ \sin(2 \times 20^\circ) = \sin(40^\circ) \]
- For \( \theta = 35^\circ \):
\[ \sin(2 \times 35^\circ) = \sin(70^\circ) \]
- For \( \theta = 60^\circ \):
\[ \sin(2 \times 60^\circ) = \sin(120^\circ) \]
- For \( \theta = 75^\circ \):
\[ \sin(2 \times 75^\circ) = \sin(150^\circ) \]

2. **Find the values of \( \sin(40^\circ) \), \( \sin(70^\circ) \), \( \sin(120^\circ) \), and \( \sin(150^\circ) \):**
- \( \sin(40^\circ) \approx 0.6428 \)
- \( \sin(70^\circ) \approx 0.9397 \)
- \( \sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) \approx 0.8660 \)
- \( \sin(150^\circ) = \sin(180^\circ - 30^\circ) = \sin(30^\circ) = 0.5 \)

3. **Compare the values of \( \sin(2\theta) \):**
- \( \sin(40^\circ) \approx 0.6428 \)
- \( \sin(70^\circ) \approx 0.9397 \) (maximum)
- \( \sin(120^\circ) \approx 0.8660 \)
- \( \sin(150^\circ) = 0.5 \)

4. **Conclusion:**
The maximum value of \( \sin(2\theta) \) occurs at \( \theta = 35^\circ \) with \( \sin(70^\circ) \approx 0.9397 \). Therefore, the projectile with the angle of projection \( 35^\circ \) will have the maximum range.

### Final Answer:
The range will be maximum for the projectile whose angle of projection is \( 35^\circ \).