Question

Where should the tank be punctured so that water coming out has maximum horizontal range?

• A. $y=\frac{h}{2}$
• B. $y=\frac{h}{3}$
• C. $y=\frac{h}{4}$
• D. None of these

Answer 1

The correct option is A

$y=\frac{h}{2}$

Here, velocity of water coming out of hole, $v=\sqrt{2g(h-y)}$
Time for water to reach ground, $t=\sqrt{\frac{2y}{g}}$
Range, $R=vt=\sqrt{2g(h-y)}\sqrt{\frac{2y}{g}}=\sqrt{4y(h-y)}$
For R to be maximum, $\frac{dR}{dy}=0\Rightarrow y=\frac{h}{2}$