Answer
The rocket travels a horizontal distance of 33.7 meters before reaching the ground.
Work Step by Step
We can find the time $t$ to reach the ground.
$y = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2y}{g}}=\sqrt{\frac{(2)(30.0~m)}{9.80~m/s^2}}$
$t = 2.474~s$
$a_x(t) = (1.60~m/s^3)~t$
$v_x(t) = v_{0x} + \int_{0}^{t}a_x(t)~dt$
$v_x(t)=v_{0x} + \int_{0}^{t}(1.60~m/s^3)~t~dt$
$v_x(t) = 12.0~m/s + (0.80~m/s^3)~t^2$
We can use $v_x(t)$ to find $x(t)$.
$x(t) = x_0 + \int_{0}^{t}v_x(t)~dt$
$x(t) = 0 + \int_{0}^{t}12.0~m/s + (0.80~m/s^3)~t^2~dt$
$x(t) = (12.0~m/s)~t + \frac{0.80~m/s^3}{3}~t^3$
When t = 2.474 s:
$x = (12.0~m/s)(2.474~s) + \frac{0.80~m/s^3}{3}(2.474~s)^3$
$x = 33.7~m$
The rocket travels a horizontal distance of 33.7 meters before reaching the ground.
