Two projectiles are fired with same initial speed from same point on ground at angles of $$(45^{\circ}-\alpha)$$ and$$ (45^{\circ}+\alpha)$$, respectively, with the horizontal direction. The ratio of their maximum heights attained is :

We need to find the ratio of maximum heights of two projectiles fired at angles $$(45° - \alpha)$$ and $$(45° + \alpha)$$ with the same initial speed. Since the maximum height of a projectile is given by $$H = \frac{u^2 \sin^2\theta}{2g}$$, we can proceed to express each height in this form.

Accordingly, the maximum heights of the two projectiles become $$H_1 = \frac{u^2 \sin^2(45° - \alpha)}{2g}$$ and $$H_2 = \frac{u^2 \sin^2(45° + \alpha)}{2g}$$.

Forming their ratio gives $$\frac{H_1}{H_2} = \frac{\sin^2(45° - \alpha)}{\sin^2(45° + \alpha)}$$.

Using the identity $$\sin(45° \mp \alpha) = \frac{\cos\alpha \mp \sin\alpha}{\sqrt{2}},$$ one obtains $$\sin^2(45° - \alpha) = \frac{(\cos\alpha - \sin\alpha)^2}{2} = \frac{1 - \sin 2\alpha}{2}$$ since $$(\cos\alpha - \sin\alpha)^2 = \cos^2\alpha - 2\sin\alpha\cos\alpha + \sin^2\alpha = 1 - \sin 2\alpha$$, and similarly $$\sin^2(45° + \alpha) = \frac{(\cos\alpha + \sin\alpha)^2}{2} = \frac{1 + \sin 2\alpha}{2}$$.

Substituting these results back into the ratio yields $$\frac{H_1}{H_2} = \frac{(1 - \sin 2\alpha)/2}{(1 + \sin 2\alpha)/2} = \frac{1 - \sin 2\alpha}{1 + \sin 2\alpha}$$.

The correct answer is Option (2): $$\frac{1 - \sin 2\alpha}{1 + \sin 2\alpha}$$.