If the angle of projection of a projector with same initial velocity exceed or fall short of `45^(@)` by equal amount `alpha`, then the ratio of horizontal rages is

Options

A. `1:2`
B. `1:3`
C. `1:4`
D. `1:1`

Explanation

To solve the problem, we need to find the ratio of horizontal ranges when the angle of projection exceeds or falls short of 45 degrees by an equal amount, α.

### Step-by-Step Solution:

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

2. **Define the Angles**:
Let:
- \( \theta_1 = 45^\circ + \alpha \) (angle exceeds 45 degrees)
- \( \theta_2 = 45^\circ - \alpha \) (angle falls short of 45 degrees)

3. **Calculate the Range for \( \theta_1 \)**:
Using the range formula for \( \theta_1 \):
\[ R_1 = \frac{u^2 \sin(2\theta_1)}{g} = \frac{u^2 \sin(2(45^\circ + \alpha))}{g} \]
Simplifying:
\[ R_1 = \frac{u^2 \sin(90^\circ + 2\alpha)}{g} = \frac{u^2 \cos(2\alpha)}{g} \]

4. **Calculate the Range for \( \theta_2 \)**:
Using the range formula for \( \theta_2 \):
\[ R_2 = \frac{u^2 \sin(2\theta_2)}{g} = \frac{u^2 \sin(2(45^\circ - \alpha))}{g} \]
Simplifying:
\[ R_2 = \frac{u^2 \sin(90^\circ - 2\alpha)}{g} = \frac{u^2 \cos(2\alpha)}{g} \]

5. **Finding the Ratio of Ranges**:
Now, we find the ratio of the ranges \( R_1 \) and \( R_2 \):
\[ \frac{R_1}{R_2} = \frac{\frac{u^2 \cos(2\alpha)}{g}}{\frac{u^2 \cos(2\alpha)}{g}} = 1 \]

6. **Conclusion**:
Therefore, the ratio of the horizontal ranges \( R_1 : R_2 \) is:
\[ R_1 : R_2 = 1 : 1 \]

### Final Answer:
The ratio of horizontal ranges is \( 1 : 1 \).

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