Chapter 8, Problem 5P

To determine

To show: The maximum height reached by a projectile as a function of its initial speed ${\nu }_0$ and its firing angle $\theta$ is $y=\frac{{{\displaystyle v}}_0^2{\sin }^2\theta }{2g}$.

Explanation of Solution

The parametric equations for path of a particle with gravity are,

$x=\left({\nu }_0\cos \theta \right)t$ (1)

And

$y=\left({\nu }_0\sin \theta \right)t-\frac{1}{2}g{t}^2$ (2)

Calculate the value of t from equation (1),

$t=\frac{x}{{\nu }_0\cos \theta }$

Substitute $\frac{x}{{\nu }_0\cos \theta }$ for t in equation (2),

$\begin{array}{c}y=\left({\nu }_0\sin \theta \right)\left(\frac{x}{{\nu }_0\cos \theta }\right)-\frac{1}{2}g{\left(\frac{x}{{\nu }_0\cos \theta }\right)}^2\\ y=\frac{x\times \sin \theta }{\cos \theta }-\frac{1}{2}g\left(\frac{x^2}{{{\displaystyle \nu }}_0^2{\cos }^2\theta }\right)\\ y=x\tan \theta -\frac{1}{2}{x}^2\left(\frac{g}{{{\displaystyle \nu }}_0^2{\cos }^2\theta }\right)\\ y=-\left(\frac{g}{2{{\displaystyle \nu }}_0^2{\cos }^2\theta }\right)x^2+x\tan \theta \end{array}$ (3)

Formula to calculate the vertex point,

$x=-\frac{b}{2a}$

Substitute $-\left(\frac{g}{2{{\displaystyle \nu }}_0^2{\cos }^2\theta }\right)$ for a, $\tan \theta$ for b in the above expression.

$\begin{array}{c}x=\frac{\tan \theta }{\frac{g}{{{\displaystyle \nu }}_0^2{\cos }^2\theta }}\\ x=\frac{\sin \theta }{\cos \theta }\cdot \frac{{{\displaystyle \nu }}_0^2{\cos }^2\theta }{g}\\ x=\frac{{{\displaystyle \nu }}_0^2\cos \theta \sin \theta }{g}\end{array}$

Substitute $\frac{{{\displaystyle \nu }}_0^2\cos \theta \sin \theta }{g}$ for x in equation (3),

$\begin{array}{c}y=-\left(\frac{g}{2{{\displaystyle \nu }}_0^2{\cos }^2\theta }\right){\left(\frac{{{\displaystyle \nu }}_0^2\cos \theta \sin \theta }{g}\right)}^2+\left(\frac{{{\displaystyle \nu }}_0^2\cos \theta \sin \theta }{g}\right)\tan \theta \\ y=-\left(\frac{g}{2{{\displaystyle \nu }}_0^2{\cos }^2\theta }\right)\left(\frac{{{\displaystyle \nu }}_0^4{\cos }^2\theta {\sin }^2\theta }{g^2}\right)+\frac{\sin \theta }{\cos \theta }\left(\frac{{{\displaystyle \nu }}_0^2\cos \theta \sin \theta }{g}\right)\\ y=-\left(\frac{1}{2}\right)\left(\frac{{{\displaystyle \nu }}_0^2{\sin }^2\theta }{g}\right)+\left(\frac{{{\displaystyle \nu }}_0^2{\sin }^2\theta }{g}\right)\\ y=\frac{1}{2}\left(\frac{{{\displaystyle \nu }}_0^2{\sin }^2\theta }{g}\right)\end{array}$

Hence, the maximum height reached by a projectile as a function of its initial speed ${\nu }_0$ and its firing angle $\theta$ is $y=\frac{{{\displaystyle \nu }}_0^2{\sin }^2\theta }{2g}$.