Chapter 4, Problem 64RE

(a)

To determine

To show: The range of the projectile is given by $R\left(\theta \right)=\frac{2v^2\cos \theta \sin \left(\theta -\alpha \right)}{g{\cos }^2\theta }$.

Answer to Problem 64RE

Explanation of Solution

Given information:

Parabola is $y=\left(\tan \theta \right)x-\frac{g}{2v^2{\cos }^2\theta }{x}^{2}0\le \theta \le \frac{\pi }{2}$ .

Calculation:

  

Consider that $y=\left(\tan \theta \right)x-\frac{g}{2v^2{\cos }^2\theta }{x}^{2}0\le \theta \le \frac{\pi }{2}$

From the figure $\tan \alpha =\frac{y}{x}$ and $y=x\tan \alpha$

Thus,

  $\begin{array}{l}x\tan \alpha =\left(\tan \theta \right)x-\frac{g}{2v^2{\cos }^2\theta }{x}^2\\ \tan \alpha =\left(\tan \theta \right)-\frac{g}{2v^2{\cos }^2\theta }x\\ x=\left(\tan \theta -\tan \alpha \right)\frac{2v^2{\cos }^2\theta }{g}\end{array}$

Next rewrite $\tan \theta =\frac{\sin \theta }{\cos \theta }$ and $\tan \alpha =\frac{\sin \alpha }{\cos \alpha }$

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

From the figure $\begin{array}{l}\cos \alpha =\frac{x}{R}\\ x=R\cos \alpha \end{array}$

So,

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

Hence $R=\frac{2v^2\cos \theta \sin \left(\theta -\alpha \right)}{g{\cos }^2\alpha }$ proved.

(b)

To determine

To find: The $\theta$ so that $R$ is maximum.

Answer to Problem 64RE

The absolute maximum of $R$ occurs when $\theta =\frac{\pi }{4}+\frac{\alpha }{2}$ .

Explanation of Solution

Given information:

The given information is

Parabola $y=\left(\tan \theta \right)x-\frac{g}{2v^2{\cos }^2\theta }{x}^{2}0\le \theta \le \frac{\pi }{2}$ .

Calculation:

To maximize $R$ , find the derivative of $R$ with respect to

Use the identity $\cos \left(A+B\right)=\cos A\cos B-\sin A\mathrm{Sin}B$

  $\begin{array}{l}R\text{'}=\frac{2v^2}{g{\cos }^2\alpha }\left[\cos \theta \cos \left(\theta -\alpha \right)-\sin \theta \sin \left(\theta -\alpha \right)\right]\\ =\frac{2v^2}{g{\cos }^2\alpha }\left[\cos \left(\theta +\left(\theta -\alpha \right)\right)\right]\\ =\frac{2v^2}{g{\cos }^2\alpha }\left[\cos \left(2\theta -\alpha \right)\right]\end{array}$

The equation $R\text{'}=0$ is satisfied when $\cos \left(2\theta -\alpha \right)=0$ that is, when

  $\begin{array}{l}\theta =\frac{\frac{\pi }{2}+\alpha }{2}\\ =\frac{\pi }{4}+\frac{\alpha }{2}\end{array}$

Thus $R\text{'}=0$ when $\theta =\frac{\pi }{4}+\frac{\alpha }{2}$ . Observe that $R\text{'}<0$ when $\theta >\frac{\pi }{4}+\frac{\alpha }{2}$ and $R\text{'}>0$ when $\theta <\frac{\pi }{4}+\frac{\alpha }{2}$ ,so by the first derivative Test.

Therefore, the absolute maximum of $R$ occurs when $\theta =\frac{\pi }{4}+\frac{\alpha }{2}$ .

(c)

To determine

To find: The range R and angle at which projectile should be fired to maximize R.

Answer to Problem 64RE

The absolute maximum of $R$ occurs when $\theta =\frac{\pi }{4}-\frac{\alpha }{2}$ .

Explanation of Solution

Given information:

The given information is Parabola $y=\left(\tan \theta \right)x-\frac{g}{2v^2{\cos }^2\theta }{x}^{2}0\le \theta \le \frac{\pi }{2}$ .

Calculation:

Draw a diagram of projectile.

  

From the figure $\tan \alpha =\frac{-y}{x}$ and $y=-x\tan \alpha$

Thus,

  $\begin{array}{l}-x\tan \alpha =\left(\tan \theta \right)x-\frac{g}{2v^2{\cos }^2\theta }{x}^2\\ -\tan \alpha =\left(\tan \theta \right)-\frac{g}{2v^2{\cos }^2\theta }x\\ x=\left(\tan \theta +\tan \alpha \right)\frac{2v^2{\cos }^2\theta }{g}\end{array}$

Next rewrite $\tan \theta =\frac{\sin \theta }{\cos \theta }$ and $\tan \alpha =\frac{\sin \alpha }{\cos \alpha }$

  $\begin{array}{l}x=\left(\frac{\sin \theta }{\cos \theta }+\frac{\sin \alpha }{\cos \alpha }\right)\frac{2v^2{\cos }^2\theta }{g}\\ x=\left(\frac{\sin \theta \cos \alpha +\cos \theta \sin \alpha }{\cos \theta \cos \alpha }\right)\frac{2v^2{\cos }^2\theta }{g}\\ x=\left(\frac{\sin \left(\theta +\alpha \right)}{\cos \alpha }\right)\frac{2v^2\cos \theta }{g}\end{array}$

From the figure $\begin{array}{l}\cos \alpha =\frac{x}{R}\\ x=R\cos \alpha \end{array}$

So,

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

To maximize, find the derivative of R with respect to

  $\begin{array}{l}R\text{'}=\frac{2v^2}{g{\cos }^2\alpha }\left[\cos \theta \cos \left(\theta +\alpha \right)-\sin \theta \sin \left(\theta +\alpha \right)\right]\\ =\frac{2v^2}{g{\cos }^2\alpha }\left[\cos \left(\theta +\left(\theta +\alpha \right)\right)\right]\\ =\frac{2v^2}{g{\cos }^2\alpha }\left[\cos \left(2\theta +\alpha \right)\right]\end{array}$

The equation $R\text{'}=0$ is satisfied when $\cos \left(2\theta +\alpha \right)=0$ that is, when

  $\begin{array}{l}\theta =\frac{\frac{\pi }{2}-\alpha }{2}\\ =\frac{\pi }{4}-\frac{\alpha }{2}\end{array}$

Thus, the absolute maximum of $R$ occurs when $\theta =\frac{\pi }{4}-\frac{\alpha }{2}$ .