Answer
(a) $3.897ft$ and $0.5625ft$
(b) $23.982ft$ and $3.462ft$
Work Step by Step
We are given $R=\frac{v_0^2sin(2\theta)}{g}$ and $H=\frac{v_0^2sin^2(\theta)}{2g}$
(a) With $v_0=12 ft/s, \theta=\frac{\pi}{6}, g=32 ft/s^2$, we have $R=\frac{v_0^2sin(2\theta)}{g}
=\frac{12^2sin2\times\frac{\pi}{6}}{32}\approx3.897ft$ and
$H=\frac{v_0^2sin^2(\theta)}{2g}=\frac{12^2sin^2\frac{\pi}{6}}{2\times32}\approx0.5625ft$
(b) With $v_0=12 ft/s, \theta=\frac{\pi}{6}, g=5.2 ft/s^2$, we have $R=\frac{v_0^2sin(2\theta)}{g}
=\frac{12^2sin2\times\frac{\pi}{6}}{5.2}\approx23.982ft$ and
$H=\frac{v_0^2sin^2(\theta)}{2g}=\frac{12^2sin^2\frac{\pi}{6}}{2\times5.2}\approx3.462ft$
