# Question #a2be2

Feb 9, 2017

C

#### Explanation:

At mean position all the potential energy has been converted in to bob's kinetic energy. Setting up the equation we get

$P E = K E$

Using formulae for both
$m g \Delta h = \frac{1}{2} m {v}^{2}$
$\implies {v}^{2} = 2 g \Delta h$
$\implies v = \sqrt{2 g \Delta h}$

Inserting given values we get
$v = \sqrt{2 \times 10 \times 0.1}$
$\implies v = \sqrt{2} {\text{ ms}}^{-} 1$