Answer
$14.9\space s$
Work Step by Step
Please see the attached image first.
We can apply equation $F=ma$ to the climber, & rock separately as follows.
$\downarrow F=ma$ ; to climber,
Let's plug known values into this equation.
$63.2g\space N-T= (63.2\space kg)\space a-(1)$
$\rightarrow F=ma$ ; to the rock
Let's plug known values into this equation.
$T=(1220\space kg)\space a-(2)$
$(1)+(2)=\gt$
$63.2\times9.8\space kgm/s^{2}= 1283.2\space kg\space a$
$0.48\space m/s^{2}= a$
Now we can apply the equation $S=ut+\frac{1}{2}at^{2}$ to the rock to find the time.
$\rightarrow S=ut+\frac{1}{2}at^{2}$
Let's plug known values into the equation.
$48.3\space m=0+\frac{1}{2}\times0.48\space m/s^{2}\times t^{2}$
$201.25\space s^{2}=t^{2}$
$t= 14.19\space s$
She has 14.19s before the rock goes over the edge
