Two particles are projected with the same velocity but at different projection angles: \(\left(\frac{\pi }{4}+\alpha \ri…

The maximum height of a projectile is given by:

\(H=\frac{{u}^{2}{\sin ⁡}^{2}\theta }{2g}\)​

where:

• \(u\) is the initial velocity,
• \(\theta\) is the angle of projection,
• \(g\) is the acceleration due to gravity.

The two angles of projection given are:

1. \({\theta }_{1}=\alpha +\frac{\pi }{4}\)​
2. \({\theta }_{2}=\alpha −\frac{\pi }{4}\)​

Using the formula for maximum height:

\({H}_{1}=\frac{{u}^{2}{\sin ⁡}^{2}(\alpha +\pi \mathrm{/}4)}{2g}\)

\({H}_{2}=\frac{{u}^{2}{\sin ⁡}^{2}(\alpha −\pi \mathrm{/}4)}{2g}\)​

The required ratio is:

\(\frac{{H}_{1}}{{H}_{2}}=\frac{{\sin ⁡}^{2}(\alpha +\pi \mathrm{/}4)}{{\sin ⁡}^{2}(\alpha −\pi \mathrm{/}4)}\)​

Using the sine angle addition and subtraction formulas:

\(\sin ⁡(\alpha +\pi \mathrm{/}4)=\frac{\sin ⁡\alpha +\cos ⁡\alpha }{\sqrt{2}}\)

​​ \(\sin ⁡(\alpha −\pi \mathrm{/}4)=\frac{\sin ⁡\alpha −\cos ⁡\alpha }{\sqrt{2}}\)​

Squaring both:

\({\sin ⁡}^{2}(\alpha +\pi \mathrm{/}4)=\frac{(\sin ⁡\alpha +\cos ⁡\alpha {)}^{2}}{2}\)

\({\sin ⁡}^{2}(\alpha −\pi \mathrm{/}4)=\frac{(\sin ⁡\alpha −\cos ⁡\alpha {)}^{2}}{2}\)​

Thus, the ratio simplifies to:

\(\frac{{H}_{1}}{{H}_{2}}=\frac{(\sin ⁡\alpha +\cos ⁡\alpha {)}^{2}}{(\sin ⁡\alpha −\cos ⁡\alpha {)}^{2}}\)​

Using the identity:

\((\sin ⁡\alpha +\cos ⁡\alpha {)}^{2}=1+\sin ⁡2\alpha\)

\((\sin ⁡\alpha −\cos ⁡\alpha {)}^{2}=1−\sin ⁡2\alpha\)

\(\frac{{H}_{1}}{{H}_{2}}=\frac{1+\sin ⁡2\alpha }{1−\sin ⁡2\alpha }\)​

Comparing with the given options, the correct answer is:

\(A\ \frac{1+\sin ⁡2\alpha }{1−\sin ⁡2\alpha }\)​