Conservation of mechanical energy

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Question

If the water falls from a dam into a turbine wheel 19.6 m below, then the velocity of water (in m/s) at the turbine is $(g = 9 .8 m / s^{2})$

Moderate

Solution

Potential energy of water = kinetic energy at turbine
$mgh = \frac{1}{2} {mv}^{2} \Rightarrow v = \sqrt{2 gh} = \sqrt{2 \times 9 .8 \times 19 .6} = 19 .6 m / s$

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