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A water barrel stands on a table of height h. If a small hole is punched in the side of the barrel at its base, it is found that the resultant stream of water strikes the ground at a horizontal distance R from the table. What is the depth of water in the barrel?

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$\frac{R^{2}}{h}$

$\frac{R^{2}}{2 h}$

$\frac{R^{2}}{4 h}$

$\frac{4 R^{2}}{h}$

answer is C.

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Detailed Solution

From Torricelli's theorem

v = $\sqrt{2 gd} - - - - - - (i)$

where y is horizontal velocity and d is the depth of water in barrel.
Time t to hit the ground is given by

$h = \frac{1}{2} {gt}^{2} or t = \sqrt{\frac{2 h}{g}}$

$∴ R = vt = \sqrt{(2 gd)} \sqrt{\frac{2 h}{g}} = 2 \sqrt{dh} (Using (i))$

$∴ R^{2} = 4 dh or d = \frac{R^{2}}{4 h}$

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