Two objects are projected with same velocity `u` however at different angles `alpha` and `beta` with the horizontal. If `alpha+beta=90^(@)`,the ratio of horizontal range of the first object to the `2` nd object will be

Options

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

Answer

To solve the problem, we need to find the ratio of the horizontal ranges of two objects projected at angles α and β, where α + β = 90°.

### Step-by-Step Solution:

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

2. **Identifying the Angles**:
Let:
- For the first object, the angle of projection is \( \alpha \).
- For the second object, the angle of projection is \( \beta \).
Given that \( \alpha + \beta = 90^\circ \), we can express \( \beta \) as:
\[ \beta = 90^\circ - \alpha \]

3. **Calculating the Ranges**:
- The range \( R_1 \) for the first object (angle \( \alpha \)):
\[ R_1 = \frac{u^2 \sin(2\alpha)}{g} \]
- The range \( R_2 \) for the second object (angle \( \beta = 90^\circ - \alpha \)):
\[ R_2 = \frac{u^2 \sin(2\beta)}{g} = \frac{u^2 \sin(2(90^\circ - \alpha))}{g} \]
Using the identity \( \sin(90^\circ - x) = \cos(x) \), we have:
\[ R_2 = \frac{u^2 \sin(180^\circ - 2\alpha)}{g} = \frac{u^2 \sin(2\alpha)}{g} \]
(since \( \sin(180^\circ - x) = \sin(x) \)).

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

5. **Conclusion**:
Therefore, the ratio of the horizontal range of the first object to the second object is:
\[ \frac{R_1}{R_2} = 1 \]