## College Physics (4th Edition)

(a) At the highest point, $v_x = v_i~cos~\theta$ At the highest point, $v_y = 0$ (b) $t = \frac{v_i~sin~\theta}{g}$ (c) $H = \frac{(v_i~sin~\theta)^2}{2g}$
(a) Since $v_x$ is constant, at the highest point, $v_x = v_i~cos~\theta$ At the highest point, $v_y = 0$ (b) At the maximum height, $v_y = 0$. We can find the time to reach the maximum height: $v_y = v_{0y} +at$ $t = \frac{v_y-v_{0y}}{a}$ $t = \frac{0-v_i~sin~\theta}{-g}$ $t = \frac{v_i~sin~\theta}{g}$ (c) We can find he maximum height $H$: $v_y^2 = v_{0y}^2+2a_yH$ $H = \frac{v_y^2-v_{0y}^2}{2a_y}$ $H = \frac{0-(v_i~sin~\theta)^2}{(2)(-g)}$ $H = \frac{(v_i~sin~\theta)^2}{2g}$