Question

if a tube have filed with water to a height 5m.and suddenly air of specific density of 0.85started to flow over it with 50m/s.find the the new height of water in tube.

Answer 1

By the law of energy conservation:

$mgh=\frac{mv^2}{2}+mgh^{'}$

$gh=\frac{v^2}{2}+gh^{'}$

$g(h-h{'})=\frac{v^2}{2}$

$h-h^{'}=\frac{v^2}{2g}$

$h^{'}=h-\frac{v^2}{2g}$

$h{'}=5-\frac{2500}{20}=-120$