Chapter 5.3, Problem 85PE

Consider an object launched from an initial height $h_0$ with an initial velocity $v_0$ at an angle $\theta$ from the horizontal. The path of the object is given by

$y=-\frac{g}{2v_0{}^2{\cos }^2\theta }{x}^2+\left(\tan \theta \right)x=h_0$

where $x$ (in ft) is the horizontal distance from the launching point $y$ (in ft) is the height above ground level, and g is the acceleration due to gravity $\left(g=32ft/{\sec }^2\text{or}9.8m/{\sec }^2\right)$ . Show that the horizontal distance traveled by a soccer ball kicked from ground level with

velocity $v_0$ at angle $\theta$ is $x=\frac{v_0{}^2\sin 2\theta }{g}.$