Answer
It takes $16.0~ms$ for the wave pulse to travel to the upper end.
Work Step by Step
We can find the linear mass density:
$\mu = \frac{m}{L}$
$\mu = \frac{W/g}{L}$
$\mu = \frac{W}{g~L}$
$\mu = \frac{0.25~N}{(9.80~m/s^2)(10.0~m)}$
$\mu = 2.55\times 10^{-3}~kg/m$
Since the weight of the string is negligible, we can let $F = 1000~N$. We can find the speed of the wave pulse:
$v = \sqrt{\frac{F}{\mu}}$
$v = \sqrt{\frac{1000~N}{2.55\times 10^{-3}~kg/m}}$
$v = 626~m/s$
We can find the time it takes the wave pulse to reach the upper end:
$t = \frac{L}{v} = \frac{10.0~m}{626~m/s} = 0.0160~s = 16.0~ms$
It takes $16.0~ms$ for the wave pulse to travel to the upper end.
