Answer
(a) $h=\frac{[v_0~sin(\theta)]^2}{2g}$
(b) At the angle $30.0^{\circ}$:
h = 14.4 m
range = 99.8 m
At the angle $45.0^{\circ}$:
h = 28.8 m
range = 115.2 m
At the angle $60.0^{\circ}$:
h = 43.2 m
range = 99.8 m
Work Step by Step
(a) $2ah=v_y^2-v_{0y}^2$
$2(-g)h=v_y^2-v_{0y}^2$
$h=\frac{v_{0y}^2-0}{2g}$
$h=\frac{[v_0~sin(\theta)]^2-0}{2g}$
$h=\frac{[v_0~sin(\theta)]^2}{2g}$
(b) Let $\theta = 30.0^{\circ}$:
$h=\frac{[v_0~sin(\theta)]^2}{2g}$
$h=\frac{[(33.6~m/s)~sin(30.0^{\circ})]^2}{(2)(9.80~m/s^2)}$
$h = 14.4~m$
$range = \frac{v_0^2~sin(2\theta)}{g}$
$range = \frac{(33.6~m/s)^2~sin(2\times 30.0)^{\circ}}{9.80~m/s^2}$
$range = 99.8~m$
Let $\theta = 45.0^{\circ}$:
$h=\frac{[v_0~sin(\theta)]^2}{2g}$
$h=\frac{[(33.6~m/s)~sin(45.0^{\circ})]^2}{(2)(9.80~m/s^2)}$
$h = 28.8~m$
$range = \frac{v_0^2~sin(2\theta)}{g}$
$range = \frac{(33.6~m/s)^2~sin(2\times 45.0)^{\circ}}{9.80~m/s^2}$
$range = 115.2~m$
Let $\theta = 60.0^{\circ}$:
$h=\frac{[v_0~sin(\theta)]^2}{2g}$
$h=\frac{[(33.6~m/s)~sin(60.0^{\circ})]^2}{(2)(9.80~m/s^2)}$
$h = 43.2~m$
$range = \frac{v_0^2~sin(2\theta)}{g}$
$range = \frac{(33.6~m/s)^2~sin(2\times 60.0)^{\circ}}{9.80~m/s^2}$
$range = 99.8~m$
