Consider a pendulum, consisting of a massless string with a mass on the end. The mass is held with the string horizontal and then released. The mass swings down and then on its way back up, the string is cut at point P when it makes an angle of θ with the vertical. What should θ be, so that the mass travels the largest horizontal distance from P by the time it returns to the height it had when the string was cut?

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Consider a pendulum, consisting of a massless string with a mass on the end. The mass is held with the string horizontal and then released. The mass swings down and then on its way back up, the string is cut at point P when it makes an angle of $θ$ with the vertical. What should $θ$ be, so that the mass travels the largest horizontal distance from P by the time it returns to the height it had when the string was cut?
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$\tan^{- 1} \sqrt{3}$

$\tan^{- 1} (\frac{1}{\sqrt{3}})$

$\tan^{- 1} (\frac{1}{\sqrt{2}})$

$\tan^{- 1} (1)$

answer is C.

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Detailed Solution

The speed angle $θ$ is given by conservation of energy
$\Rightarrow \frac{1}{2} m v^{2} = m g h \Rightarrow \frac{1}{2} m v^{2} = m g l \cos θ \Rightarrow v = \sqrt{2 g l \cos θ}$
So we have the following setup of projectile motion.
$v = \sqrt{2 g l \cos θ}$
$v_{x} = v \cos θ$
$v_{y} = v \sin θ$
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$R = \frac{2 v^{2} \sin θ \cos θ}{g} = \frac{2 (2 g l \cos θ) \sin θ \cos θ}{g}$ $= 4 l \cos^{2} θ \sin θ$
Maximize this $\Rightarrow$ take the derivative and equate it to zero
$\frac{d R}{d θ} = - 8 l \cos θ \sin^{2} θ + 4 l \cos^{3} θ = 0 \Rightarrow 2 \sin^{2} θ = \cos^{2} θ \Rightarrow \tan^{2} θ = \frac{1}{2} \Rightarrow \tan θ = \frac{1}{\sqrt{2}}$