## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

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.