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

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

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$y = \frac{h}{3}$

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$y = \frac{h}{4}$

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None of these

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Solution

The correct option is A
$y = \frac{h}{2}$

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

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

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

The correct option is A y=h2 Here, velocity of water coming out of hole, v=√2g(h−y) Time for water to reach ground, t=√2yg Range, R=vt=√2g(h−y)√2yg=√4y(h−y) For R to be maximum, dRdy=0⇒y=h2