Physics - Fluid Question with Solution | TestHub

Let the speed of fluid at outlet is$v$.

Applying Bernoulli's theorem,

$p_1+\rho gh+\frac{1}{2}\rho v^2=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{ }$

$p_0+\rho gh+0=p_0+0+\frac{1}{2}\rho v^2$

$\therefore v=\sqrt{2gh}$...(i)

At outlet,



$v\sin \theta$vertical velocity of the fluid at outlet,

$v^2=u^2+2g{h}_1$

For maximum height,

$0={\left(v\sin \theta \right)}^2-2g{h}_1$

$\therefore h_1=\frac{v^2{\sin }^2\theta }{2g}$

$=\frac{2gh{\sin }^2\theta }{2g}$[from Equation (i)]

$=h{\sin }^2\theta$

$h_1=h{\sin }^2\theta$