Solution:

Given,

Angle of projection of first projectile, $ =\theta $

Angle of projection of second projectile, $ =(90-\theta)$

Velocity of projection, $ u=78.4 \mathrm{~ms}^{-1} \\ $

and, $ \mathrm{H}_{2}=\mathrm{H}_{1}+36.4 \mathrm{~m} \\ $

$ \mathrm{H}_{2}-\mathrm{H}_{1}=36.4 \mathrm{~m} \\ $

Where, $\mathrm{H}_{1}$ is the maximum height of first one and $\mathrm{H}_{2}$ is the maximum height of second one.

$ \mathrm{H}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}} \\ $ and , $ \mathrm{H}_{2}=\frac{\mathrm{u}^{2} \sin ^{2}(90-\theta)}{2 \mathrm{~g}}=\frac{\mathrm{u}^{2} \cos ^{2} \theta}{2 \mathrm{~g}} \\ $

Adding, $$ \mathrm{H}_{2}+\mathrm{H}_{1}=\frac{\mathrm{u}^{2} \cos ^{2} \theta}{2 \mathrm{~g}}+\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\\ $$

$$ \begin{aligned} &=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}}=\frac{78.4 \times 78.4}{2 \times 9.8}=313.6 \\\\ 2 \mathrm{H}_{2} &=36.4+313.6=350 \\\\ \mathrm{H}_{2} &=175 \mathrm{~m} \\\\ \end{aligned} $$

Also, $$ \mathrm{H}_{1}=\mathrm{H}_{2}-36.4=175-36.4 \\ $$

$$ \mathrm{H}=138.6 \mathrm{~m}\\ $$