JEE Physics 2023 Question with Solution

Given: Two projectiles are projected with the same initial speed at angles $30^\circ$ and $60^\circ$.

Find: The ratio of their maximum heights.

In projectile motion, the maximum height is given by

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

For $\theta = 30^\circ$,

$$H_1 = \frac{u^2 \sin^2 30^\circ}{2g} = \frac{u^2 \left(\frac{1}{2}\right)^2}{2g} = \frac{u^2}{8g}$$

For $\theta = 60^\circ$,

$$H_2 = \frac{u^2 \sin^2 60^\circ}{2g} = \frac{u^2 \left(\frac{\sqrt{3}}{2}\right)^2}{2g} = \frac{3u^2}{8g}$$

Now,

$$\frac{H_1}{H_2} = \frac{\frac{u^2}{8g}}{\frac{3u^2}{8g}} = \frac{1}{3}$$

Therefore, the ratio of maximum heights is $1 : 3$. The solution working gives the correct result, so the correct option should be D. The provided solution labels B, which is inconsistent with the working and the listed options.