A pendulum of length 0.471 m,with a bob of mass 0.593 kg is moved out to an angle and released from rest.The speed of the bob at the bottom of the pendulum oscillation is 1.20 m/s.What is the angle from which the pendulum is released? (in degrees)

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The correct answer is A.The speed of the bob at the bottom of the pendulum oscillation is given by:$$v = \sqrt{2gl(1 - \cos 	heta)}$$where:* v is the speed of the bob (1.20 m/s)* g is the acceleration due to gravity (9.81 m/s^2)* l is the length of the pendulum (0.471 m)* θ is the angle from which the pendulum is releasedSolving for θ, we get:$$	heta = \cos^{-1}\left(1 - \frac{v^2}{2gl}ight)$$$$	heta = \cos^{-1}\left(1 - \frac{(1.20 	ext{ m/s})^2}{2(9.81 	ext{ m/s}^2)(0.471 	ext{ m})}ight)$$$$	heta = 32.4^\circ$$

A Pendulum of Length 0