1. Formula for horizontal range

For a projectile launched with speed $u$ at angle $\theta$ to the horizontal, the horizontal range is

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

2. Range of the first object

If the first object is projected at angle $\alpha$, then

$R_1 = \frac{u^2 \sin 2\alpha}{g}$

3. Range of the second object

If the second object is projected at angle $\beta$, then

$R_2 = \frac{u^2 \sin 2\beta}{g}$

Given:

$\alpha + \beta = 90^\circ \implies \beta = 90^\circ - \alpha$

So,

$2\beta = 180^\circ - 2\alpha$

Hence,

$\sin 2\beta = \sin(180^\circ - 2\alpha) = \sin 2\alpha$

Therefore,

$R_2 = \frac{u^2 \sin 2\alpha}{g}$

4. Take the ratio

$\frac{R_1}{R_2} = \frac{\sin 2\alpha}{\sin 2\beta} = 1$

So the ratio is

$R_1 : R_2 = 1 : 1$

5. Option check

• A: $1:1$ ✅
• B: $2:1$ ❌
• C: $1:2$ ❌
• D: $4:1$ ❌

Therefore, the correct answer is A.