QJEEM23Ph7a4f: JEE Main 2023 Physics - Class 11 JEEM23Ph7a4f Solution: Two bodies are projected from ground with same speeds at two…

When two bodies are projected from the ground at the same speed of $40 \mathrm{~ms}^{-1}$ but at different angles, and they achieve the same range, we can derive the following:

Given that one projectile is launched at an angle of $60^\circ$ with respect to the horizontal, let's denote the angles of projection as $\theta_1$ and $\theta_2$ . For the ranges to be equal, we know:

$\theta_1 + \theta_2 = 90^\circ$

Since $\theta_1 = 60^\circ$ , we can find $\theta_2$ as:

$\theta_2 = 30^\circ$

Next, to find the sum of the maximum heights attained by both projectiles, we use the formula for the maximum height $H_{\max}$ :

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

For the first body projected at $60^\circ$ :

$\left(H_{\max}\right)_1 = \frac{(40)^2 \sin^2 60^\circ}{2 \times 10}$

Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$ :

$\left(H_{\max}\right)_1 = \frac{1600 \times \left(\frac{\sqrt{3}}{2}\right)^2}{20} = \frac{1600 \times \frac{3}{4}}{20} = \frac{1200}{20} = 60 \mathrm{~m}$

For the second body projected at $30^\circ$ :

$\left(H_{\max}\right)_2 = \frac{(40)^2 \sin^2 30^\circ}{2 \times 10}$

Since $\sin 30^\circ = \frac{1}{2}$ :

$\left(H_{\max}\right)_2 = \frac{1600 \times \left(\frac{1}{2}\right)^2}{20} = \frac{1600 \times \frac{1}{4}}{20} = \frac{400}{20} = 20 \mathrm{~m}$

Therefore, the sum of the maximum heights attained by both projectiles is:

$\left(H_{\max}\right)_1 + \left(H_{\max}\right)_2 = 60 \mathrm{~m} + 20 \mathrm{~m} = 80 \mathrm{~m}$