The bob of a pendulum was released from a horizontal position. The...

Given the length of the pendulum is $ℓ = 10m$.

The initial potential energy of the pendulum at the horizontal position is given by

$U = mgℓ . . . \left(1\right)$

The final kinetic energy of the pendulum at the lowermost position is given by

$K = \dfrac{1}{2}m{v}^{2} . . . \left(2\right)$

So, from equations (1) and (2), it follows that

$\dfrac{9}{10}mgℓ = \dfrac{1}{2}m{v}^{2} ⇒ \dfrac{9}{10} \times 10 \times 10 = \dfrac{1}{2}{v}^{2} ⇒ {v}^{2} = 180 ⇒ v = \sqrt{180} = 6\sqrt{5}m{s}^{- 1}$