Problem 12.39

12.39. a)
The period is given by
$T = 2\pi \sqrt{\frac{\ell}{g}}$
or T = 3.00 sec

b)
The energy is given by
$E = \frac{1}{2}m v^{2}$
$E = \frac{1}{2} (6.74)(2.06)^{2} ~=~ 14.3 J$

c)
At maximum angular displacement,
$mgh = \frac{1}{2} m v^{2}$
so, the height is $h = \frac{v^{2}}{2g} ~=~ .217 m$
From the geometry of the problem, $h = L - L \cos \theta$
$\cos \theta = 1 - \frac{h}{L}$
or $\theta = 25.5{}^{\circ}$