Answer
(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) $H = \frac{(v_i~sin~\theta)^2}{2g}$
Work Step by Step
(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 $H$:
$v_y^2 = v_{0y}^2+2a_yH$
$H = \frac{v_y^2-v_{0y}^2}{2a_y}$
$H = \frac{0-(v_i~sin~\theta)^2}{(2)(-g)}$
$H = \frac{(v_i~sin~\theta)^2}{2g}$
