Keeping the velocity of projection constant, the angle of projection is increased from 0° to 90°. Then the horizontal range of the projectile

Options

A. goes on increasing up to 90°
B. decreases up to 90°
C. increases up to 45° and decreases afterwards
D. decreases up to 45° and increases afterwards

Answer

To solve the problem, we need to analyze how the horizontal range of a projectile changes as the angle of projection increases from 0° to 90°, while keeping the velocity of projection constant.

### Step-by-Step Solution:

1. **Understanding the Formula for Horizontal Range**:
The horizontal range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where:
- \( R \) is the horizontal range,
- \( u \) is the initial velocity of projection,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

2. **Analyzing the Sine Function**:
The term \( \sin 2\theta \) is crucial in determining the range. The sine function has specific properties:
- \( \sin 2\theta \) increases from 0 to 1 as \( 2\theta \) goes from 0° to 90°.
- This means \( \sin 2\theta \) reaches its maximum value of 1 when \( 2\theta = 90° \) or \( \theta = 45° \).

3. **Behavior of the Range with Respect to Angle**:
- As \( \theta \) increases from 0° to 45°, \( \sin 2\theta \) increases, and thus the range \( R \) increases.
- At \( \theta = 45° \), the range \( R \) reaches its maximum value.
- As \( \theta \) continues to increase from 45° to 90°, \( \sin 2\theta \) starts to decrease, which means the range \( R \) will also decrease.

4. **Conclusion**:
Therefore, the horizontal range of the projectile increases up to an angle of 45° and then decreases after that. This leads us to the conclusion that the correct option is:
- **Option 3**: The range increases up to 45° and decreases afterwards.