Chapter 14.2, Problem 49PPS

To determine

To find :The expression in terms of $\tan \theta$

Answer to Problem 49PPS

1

Explanation of Solution

Given information :

We know that, $y=\frac{-g{x}^2}{2V_0^2{\cos }^2\theta }+\frac{x\sin \theta }{\cos \theta }$

Formula Used :

  $R=\frac{V_0^2\sin 2\theta }{g}$

Calculation :

  $\begin{array}{l}y=\frac{-g{x}^2}{2v_0^2{\cos }^2\theta }+\frac{x\sin \theta }{\cos \theta }\\ y=x\tan \theta -\frac{g{x}^2}{2v_0^2{\cos }^2\theta }\\ y=x\tan \theta \left[1-\frac{gx}{2v_0^2{\cos }^2\theta }\frac{1}{\tan \theta }\right]\\ y=x\tan \theta \left[1-\frac{x}{\frac{2v_0^2{\cos }^2\theta }{g}}\cdot \frac{\sin \theta }{\cos \theta }\right]\\ y=x\tan \theta \left[1-\frac{x}{\frac{2v_0^2\cos \theta \cdot \sin \theta }{g}}\right]\\ y=x\tan \theta \left[1-\frac{x}{R}\right]\end{array}$