A pendulum bob has a speed of $3\, m/s$ at its lowest position. The pendulum is $0.5\, m$ long. The speed of the bob, when the length makes an angle of ${60^o}$ to the vertical, will be ..... $m/s$ (If $g = 10\,m/{s^2}$)
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(d) Let bob velocity be $v$ at point $B$ where it makes an angle of $60^o$ with the vertical, then using conservation of mechanical energy
$K{E_A} + P{E_A} = K{E_B} + P{E_B}$
==> $\frac{1}{2}m \times {3^2} = \frac{1}{2}m{v^2} + mgl(1 - \cos \theta )$
$⇒ 9 = {v^2} + 2 \times 10 \times 0.5 \times \frac{1}{2} $
$⇒ v = 2\,m/s$
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