Two projectiles thrown at $$30°$$ and $$45°$$ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is

We need to find the ratio of initial velocities of two projectiles thrown at $$30°$$ and $$45°$$ with the horizontal, given that they reach maximum height in the same time.

We start by recalling that for a projectile launched with initial velocity $$v$$ at angle $$\theta$$, the time to reach maximum height is: $$t = \frac{v \sin\theta}{g}$$

Next, let the initial velocities be $$v_1$$ (at $$30°$$) and $$v_2$$ (at $$45°$$). Since both reach maximum height in the same time, we have: $$\frac{v_1 \sin 30°}{g} = \frac{v_2 \sin 45°}{g}$$

This simplifies to $$v_1 \sin 30° = v_2 \sin 45°$$

$$v_1 \times \frac{1}{2} = v_2 \times \frac{\sqrt{2}}{2}$$

Therefore, $$\frac{v_1}{v_2} = \frac{\sqrt{2}/2}{1/2} = \frac{\sqrt{2}}{1} = \sqrt{2}$$

Hence, $$v_1 : v_2 = \sqrt{2} : 1$$.

The correct answer is Option C: $$\sqrt{2} : 1$$.