Chapter 7.4, Problem 38E

To determine

To prove: $R=\frac{v^2\sqrt{2}}{g}\left(\sin 2\theta -\cos 2\theta -1\right)$ by using the given information.

Answer to Problem 38E

See explanation

Explanation of Solution

Given information:

  $\begin{array}{l}R=\frac{2v^2\cos \theta \sin \left(\theta -\alpha \right)}{g{\cos }^2\alpha }\\ \alpha =45°\end{array}$

Calculation:

Substitute the value of $\alpha$ in the given equation and follow the steps mentioned below.

  $\begin{array}{c}R=\frac{2v^2\cos \theta \sin \left(\theta -45°\right)}{g{\cos }^{2}45°}\\ =\frac{2v^2\cos \theta \left(\sin \theta \cos 45°-\cos \theta \sin 45°\right)}{g{\cos }^{2}45°}\left[\text{Use the differece identity of sine}\right]\\ =\frac{2v^2\cos \theta \left(\sin \theta \left(\raisebox{1ex}{$\sqrt{2}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)-\cos \theta \left(\raisebox{1ex}{$\sqrt{2}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\right)}{g{\left(\raisebox{1ex}{$\sqrt{2}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2}\\ =\frac{2\left(\raisebox{1ex}{$\sqrt{2}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)v^2\cos \theta \left(\sin \theta -\cos \theta \right)}{g\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}\\ =\frac{\sqrt{2}{v}^2\left(2\sin \theta \cos \theta -2{\cos }^2\theta \right)}{g}\\ =\frac{\sqrt{2}{v}^2\left(\sin 2\theta -\left(\cos 2\theta +1\right)\right)}{g}\\ =\frac{v^2\sqrt{2}}{g}\left(\sin 2\theta -\cos 2\theta -1\right)\end{array}$

Thus, the given equation is proved.