Two bodies are projected from ground with same speeds 40m s ^ - 1...

For two projectiles with same speed to have the same horizontal range, angle of projection will be ${\theta}_{1} + {\theta}_{2} = 90 °$.

Therefore, ${\theta}_{2} = 90 ° - 60 ° = 30 °$.

Now, required sum of maximum heights will be,

{\left({H}_\max\right)}_{1} + {\left({H}_\max\right)}_{2} = \dfrac{{u}^{2}\sin^{2}{\theta}_{1}}{2g} + \dfrac{{u}^{2}\sin^{2}{\theta}_{2}}{2g}

$= \dfrac{{40}^{2}}{20}\left(\dfrac{1}{4} + \dfrac{3}{4}\right) = 80m$