JEE Physics 2025 Question with Solution

Given: Two projectiles are projected with the same initial speed at angles $\theta_1 = 45^\circ + \alpha$ and $\theta_2 = 45^\circ - \alpha$.

Find: The ratio $\frac{T_1}{T_2}$ of their times of flight.

For a projectile, time of flight is

$$T = \frac{2u \sin\theta}{g}$$

So,

$$T_1 = \frac{2u\sin(45^\circ + \alpha)}{g}$$

and

$$T_2 = \frac{2u\sin(45^\circ - \alpha)}{g}$$

Therefore,

$$\frac{T_1}{T_2} = \frac{\sin(45^\circ + \alpha)}{\sin(45^\circ - \alpha)}$$

Using sine addition and subtraction identities,

$$\sin(45^\circ + \alpha) = \sin45^\circ\cos\alpha + \cos45^\circ\sin\alpha$$ $$\sin(45^\circ - \alpha) = \sin45^\circ\cos\alpha - \cos45^\circ\sin\alpha$$

Since $\sin45^\circ = \cos45^\circ = \frac{1}{\sqrt{2}}$,

$$\frac{T_1}{T_2} = \frac{\frac{1}{\sqrt{2}}\cos\alpha + \frac{1}{\sqrt{2}}\sin\alpha}{\frac{1}{\sqrt{2}}\cos\alpha - \frac{1}{\sqrt{2}}\sin\alpha}$$ $$\frac{T_1}{T_2} = \frac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha}$$

Dividing numerator and denominator by $\cos\alpha$,

$$\frac{T_1}{T_2} = \frac{1 + \tan\alpha}{1 - \tan\alpha}$$

Therefore, the correct option is D.

Given: $\theta_1 = 45^\circ + \alpha$ and $\theta_2 = 45^\circ - \alpha$ with the same initial speed.

Find: $\frac{T_1}{T_2}$.

Because time of flight is proportional to $\sin\theta$,

$$\frac{T_1}{T_2} = \frac{\sin(45^\circ + \alpha)}{\sin(45^\circ - \alpha)}$$

Now use

$$\sin(45^\circ + \alpha) = \frac{\cos\alpha + \sin\alpha}{\sqrt{2}}$$

and

$$\sin(45^\circ - \alpha) = \frac{\cos\alpha - \sin\alpha}{\sqrt{2}}$$

Hence,

$$\frac{T_1}{T_2} = \frac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha} = \frac{1 + \tan\alpha}{1 - \tan\alpha}$$

This works quickly because the common factors $\frac{2u}{g}$ and $\frac{1}{\sqrt{2}}$ cancel immediately. Therefore, the correct option is D.