Two bodies are projected from ground with same speeds 40   m   s - 1 at two different angles with…

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,

${\left(H_{\max }\right)}_1+{\left(H_{\max }\right)}_2=\frac{u^2{\sin }^2{\theta }_1}{2g}+\frac{u^2{\sin }^2{\theta }_2}{2g}$

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