## Precalculus (10th Edition)

$R\approx19541.95$ meters $H\approx2278.14$ meters
Substitute $30^\circ$ to $\theta$ and $500$ to $v_o$ to obtain: $R=\frac{v_0^2\sin{(2\theta)}}{g}=\frac{500^2\sin{(2\cdot25^\circ)}}{9.8}=\frac{250000\cdot0.766}{9.8}\approx19541.95$ meters $H=\frac{v_0^2\sin^2{(\theta)}}{2g}=\frac{500^2\sin^2{(25^\circ)}}{2\cdot9.8}=\frac{250000\cdot(0.423)^2}{2\cdot9.8}\approx2278.14$ meters