## University Physics with Modern Physics (14th Edition)

(a) We can find $v$ at the top of the incline. $v^2 = v_0^2+2ad = 0 + 2ad$ $v = \sqrt{2ad} = \sqrt{(2)(1.90~m/s^2)(200.0~m)}$ $v = 27.57~m/s$ We can find $v_y$ at the top of the incline. $v_y = (27.57~m/s)~sin(35.0^{\circ}) = 15.81~m/s$ We can use 15.81 m/s as $v_{0y}$ in the next part of the question. We can find the maximum height $y$ above the top of the incline. $y = \frac{v_y^2-v_{0y}^2}{2g} = \frac{0-(15.81~m/s)^2}{(2)(-9.80~m/s^2)}$ $y = 12.8~m$ Let $h$ be the height at the top of the incline. $h = (200.0~m)~sin(35.0^{\circ}) = 114.7~m$ The maximum height above the ground is 12.8 m + 114.7 m which is 127.5 meters. (b) We can find $v_y$ when the rocket returns to the ground. $v_y^2 = 0+2gy = 2gy$ $v_y = \sqrt{2gy} = \sqrt{(2)(9.80~m/s^2)(127.5~m)}$ $v_y = 50.0~m/s$ (toward the ground) We can find the time $t$ from the time to leave the top of the incline to return to the ground. $t = \frac{v_y - v_{0y}}{g} = \frac{(-50.0~m/s)-(15.81~m/s)}{-9.80~m/s^2}$ $t = 6.72~s$ We can find the horizontal distance $x$ from the top of the incline. $x = v_x~t = (27.57~m/s)~cos(35.0^{\circ})(6.72~s)$ $x = 151.8~m$ Let $L$ be the length of the incline. $L = (200.0~m)~cos(35.0^{\circ}) = 163.8~m$ The horizontal range beyond point A is 151.8 m + 163.8 m which is 315.6 meters.