Question

A tank is filled with water upto a height $H$. Water is allowed to come out of a hole $P$ in one of the walls at a depth $h$ below the surface of water (see figure). Express the horizontal distance $X$ in terms of $H$ and $h$.

• A. $X=\sqrt{h(H-h)}$
• B. $x=\sqrt{\frac{h}{2}(H-h)}$
• C. $X=2\sqrt{h(H-h)}$
• D. $X=4\sqrt{h(H-h)}$

Answer 1

The correct option is C $X=2\sqrt{h(H-h)}$

Along vertical -
Vertical distance covered by water before striking the ground $=(H-h)$
On applying $s=ut-\frac{1}{2}g{t}^2$ as motion of water along the vertical is uniformly accelerated motion under gravity.
$(H-h)=\frac{1}{2}g{t}^2$
$\Rightarrow t=\sqrt{\frac{2(H-h)}{g}}$ .........(1)


Along horizontal -
Horizontal velocity of water coming out of hole at $P$, $v=\sqrt{2gh}$
$\therefore$ Horizontal range, $X=vt$
$=\sqrt{2gh}\times \sqrt{\frac{2(H-h)}{g}}$ $[$ from (1) $]$
$\Rightarrow X=2\sqrt{h(H-h)}$