JEE Physics 2023 Question with Solution

Given: Two projectiles are projected with the same velocity $u$ at angles $\alpha$ and $\beta$ such that $\alpha + \beta = 90^\circ$.

Find: The ratio of horizontal ranges $R_1:R_2$.

For a projectile, the horizontal range is

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

Range for projection angle $\alpha$ is

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

Range for projection angle $\beta$ is

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

Given,

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

Therefore,

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

Substituting in $R_2$,

$$R_2 = \frac{u^2 \sin 2(90^\circ-\alpha)}{g}$$ $$R_2 = \frac{u^2 \sin (180^\circ-2\alpha)}{g}$$

Using $\sin(180^\circ-\theta)=\sin\theta$,

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

Hence,

$$R_2 = R_1$$

So,

$$\frac{R_1}{R_2} = \frac{1}{1}$$

Therefore, the ratio of horizontal ranges is $1:1$ and the correct option is D.

Given: The two projection angles are complementary, so $\alpha + \beta = 90^\circ$.

Find: The ratio of ranges.

For the same speed $u$, range depends on $\sin 2\theta$. If $\beta = 90^\circ - \alpha$, then

$$\sin 2\beta = \sin 2(90^\circ-\alpha) = \sin(180^\circ-2\alpha) = \sin 2\alpha$$

Hence both projectiles have the same range.

Therefore, $R_1:R_2 = 1:1$, so the correct option is D.