Answer
It takes $~~4.3~s~~$ to reach the end of the string.
Work Step by Step
In part (a), we found that the magnitude of the linear acceleration is $~~0.13~m/s^2$
We can find the time it takes to reach the end of the string:
$y = \frac{1}{2}at^2$
$t = \sqrt{\frac{2y}{a}}$
$t = \sqrt{\frac{(2)(120~cm)}{13~cm/s^2}}$
$t = 4.3~s$
It takes $~~4.3~s~~$ to reach the end of the string.
