Two bodies are projected from ground with same speeds $40 \mathrm{~ms}^{-1}$ at two different angles with respect to horizontal. The bodies were found to have same range. If one of the body was projected at an angle of $60^{\circ}$, with horizontal then sum of the maximum heights, attained by the two projectiles, is $\mathrm{m}$. (Given $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )

<p>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:</p> <p>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:</p> <p>$ \theta_1 + \theta_2 = 90^\circ $</p> <p>Since $\theta_1 = 60^\circ$, we can find $\theta_2$ as:</p> <p>$ \theta_2 = 30^\circ $</p> <p>Next, to find the sum of the maximum heights attained by both projectiles, we use the formula for the maximum height $H_{\max}$:</p> <p>$ H_{\max} = \frac{u^2 \sin^2 \theta}{2g} $</p> <p>For the first body projected at $60^\circ$:</p> <p>$ \left(H_{\max}\right)_1 = \frac{(40)^2 \sin^2 60^\circ}{2 \times 10} $</p> <p>Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$:</p> <p>$ \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} $</p> <p>For the second body projected at $30^\circ$:</p> <p>$ \left(H_{\max}\right)_2 = \frac{(40)^2 \sin^2 30^\circ}{2 \times 10} $</p> <p>Since $\sin 30^\circ = \frac{1}{2}$:</p> <p>$ \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} $</p> <p>Therefore, the sum of the maximum heights attained by both projectiles is:</p> <p>$ \left(H_{\max}\right)_1 + \left(H_{\max}\right)_2 = 60 \mathrm{~m} + 20 \mathrm{~m} = 80 \mathrm{~m} $</p>