Answer
$\theta = 28.5^{\circ}$
Work Step by Step
In part (a), we found that the maximum height is $~~h_1 = 0.19~m$
In part (b), we found that the maximum speed is $v = 1.9~m/s$
One third of this value is $\frac{1.9~m/s}{3} = 0.63~m/s$
We can use conservation of energy to find the height $h_2$:
$K_2+U_2 = K_1+U_1$
$U_2 = 0+U_1-K_2$
$mgh_2 = mgh_1-\frac{1}{2}mv_2^2$
$h_2 = h_1-\frac{v_2^2}{2g}$
$h_2 = (0.19~m)-\frac{(0.63~m/s)^2}{(2)(9.8~m/s^2)}$
$h_2 = 0.17~m$
We can find the angle $\theta$ the string makes with the vertical:
$cos~\theta = \frac{L-h_2}{L}$
$\theta = cos^{-1}~(\frac{L-h_2}{L})$
$\theta = cos^{-1}~(\frac{1.4~m-0.17~m}{1.4~m})$
$\theta = 28.5^{\circ}$
