Answer
It takes $~~0.88~s~~$ to reach the end of the string.
Work Step by Step
We can use Equation (11-10) to find the magnitude of the linear acceleration:
$a = \frac{g~sin~\theta}{1+I/MR^2}$
$a = \frac{(9.8~m/s^2)~sin~90^{\circ}}{1+\frac{950~g~cm^2}{(120~g)(0.32~cm)^2}}$
$a = 0.13~m/s^2$
We can find the time it takes to reach the end of the string:
$y = v_0~t+\frac{1}{2}at^2$
$\frac{1}{2}at^2+v_0~t-y = 0$
$\frac{1}{2}(0.13~m/s^2)t^2+(1.3~m/s)~t-1.20~m = 0$
$(0.065~m/s^2)t^2+(1.3~m/s)~t-1.20~m = 0$
We can use the quadratic formula to find $t$:
$t = \frac{-1.3 \pm \sqrt{(1.3)^2-(4)(0.065)(-1.20)}}{(2)(0.065)}$
$t = \frac{-1.3 \pm \sqrt{2.0}}{0.13}$
$t = -21, 0.88~s$
Since time $t$ is positive, the solution is $t = 0.88$
It takes $~~0.88~s~~$ to reach the end of the string.
