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:

The horizontal range of a projectile is:

$$R = \frac{u^2 \sin 2\theta}{g}$$

For the first object (angle $$\alpha$$): $$R_1 = \frac{u^2 \sin 2\alpha}{g}$$

For the second object (angle $$\beta$$): $$R_2 = \frac{u^2 \sin 2\beta}{g}$$

Since $$\alpha + \beta = 90°$$, we have $$\beta = 90° - \alpha$$.

$$\sin 2\beta = \sin 2(90° - \alpha) = \sin(180° - 2\alpha) = \sin 2\alpha$$

Therefore $$R_1 = R_2$$, and the ratio is $$1 : 1$$.

The correct answer is Option 4: 1 : 1.