Answer
(a) $v = \sqrt{(2h)(\frac{F_{thrust} - mg}{m})}$
(b) $v = 54.3~m/s$
Work Step by Step
(a) We can find an expression for the acceleration of the rocket.
$\sum F = ma$
$F_{thrust} - mg = ma$
$a = \frac{F_{thrust} - mg}{m}$
We can find the speed when the rocket reaches a height $h$.
$v^2 = 2ah$
$v = \sqrt{2ah}$
$v = \sqrt{(2h)(\frac{F_{thrust} - mg}{m})}$
(b) We can use the expression in part (a) to find the speed of the rocket.
$v = \sqrt{(2h)(\frac{F_{thrust} - mg}{m})}$
$v = \sqrt{(2)(85~m)[\frac{9.5~N - (0.350~kg)(9.80~m/s^2}{0.350~kg}]}$
$v = 54.3~m/s$
