Physics - Fluid Question with Solution | TestHub

By equation of continuity,

$$\begin{array}{rlrlrlrl}& A_1{V}_1=A_2{V}_2& S_1\times u=S_2\times u\cos \theta & \Rightarrow \theta ={\cos }^{-1}\left(\frac{S_1}{S_2}\right)& \therefore \sin \theta =\sqrt{1-\frac{S_1^2}{S_2^2}}& \text{ Now range }R=\frac{u^2\sin 2\theta }{g}=\frac{u^{2}2\sin \theta \cos \theta }{g}& =\frac{2u^2}{g}\frac{S_1}{S_2}\sqrt{1-\frac{S_1^2}{S_2^2}}& H_{\max }=\frac{u^2{\sin }^2\theta }{2g}=\frac{u^2}{2g}\left(1-\frac{S_1^2}{S_2^2}\right)\end{array}$$

Rate of flow volume is $\left({\mathrm{S}}_1\times \mathrm{u}\right)$ or $\left({\mathrm{S}}_2\times \mathrm{u}\cos \theta \right)$ etc.