Answer to Question #163132 in Physics for ajay

Question

A hole is made on the wall of cylindrical vessel, containing a liquid upto a height h. Water is coming out through the hole and falls at a distance ‘t’ from the base of the cylinder. t is found to be maximum, when hole is created at depth of 2xh. Find the value of x?

Answer 1

Horizontal velocity of water out of hole,

$$u=\sqrt{2g(2xh)}=2\sqrt{gxh}$$

Height of hole from ground level $h-2xh$. The time taken buy water to cover vertical distance will be

$$(1-2x)h=0.5gs^2\\s=\sqrt{\frac{2(1-2x)h}{g}}$$

Range is

$$t=2\sqrt{gxh}\sqrt{\frac{2(1-2x)h}{g}}=2h\sqrt{2x(1-2x)}$$

Horizontal range will be maximum if

$$x=\frac{1}{4}$$