#### Answer

The rocket is 111 meters from the launch pad.

#### Work Step by Step

We can find the horizontal displacement after 6.00 seconds.
$x = \frac{1}{2}a_xt^2$
$x = \frac{1}{2}(1.50~m/s^2)(6.00~s)^2$
$x = 27.0~m$
We can find the vertical displacement after 6.00 seconds.
$y = \frac{1}{2}a_yt^2$
$y = \frac{1}{2}(6.00~m/s^2)(6.00~s)^2$
$y = 108~m$
We can find the distance from the launch pad.
$d = \sqrt{(x)^2+(y)^2}$
$d = \sqrt{(27.0~m)^2+(108~m)^2}$
$d = 111~m$
The rocket is 111 meters from the launch pad.