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Q.

The simple pendulum bob is given a horizontal velocity $\sqrt{g l}$ at the bottom. The angle with the vertical through which the ball swings before its velocity becomes zero is

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a

120 °

b

30 °

c

60 °

d

450 °

answer is B.

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Detailed Solution

v^{2} = v_{0}^{2} - 2 g l (1 - \cos θ)

$v = 0$

$0 = g l - 2 g l (1 - \cos θ)$

$8 l = 2 g l (1 - \cos θ)$

$2 = 1 - \cos θ$

$\cos θ = \frac{1}{2}$

$θ = 60 °$

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