Answer
The rocket should start a distance of 142 meters from the base.
Work Step by Step
$K_1+U_1+\sum W = K_2+U_2$
$0 + mgh + F~d - F_f~d = \frac{1}{2}mv^2+0$
$mgd~sin(\theta) + F~d - F_f~d = \frac{1}{2}mv^2$
$d= \frac{mv^2}{2mg~sin(\theta) + 2F - 2F_f}$
$d= \frac{(1500~kg)(50.0~m/s)^2}{(2)(1500~kg)(9.80~m/s^2)~sin(53^{\circ}) + (2)(2000~N) - (2)(500~N)}$
$d = 142~m$
The rocket should start a distance of 142 meters from the base.
