Chapter 4.8, Problem 31P

To determine

To prove: The greatest height of the projectile is $\frac{v^2}{2g}$ .

Explanation of Solution

Given information: Equation, $h=vt-\frac{1}{2}g{t}^2$ . Where $v$ is initial speed $g$ is the gravitational constant.

Proof: The given equation is, $h=vt-\frac{1}{2}g{t}^2$

Rearranged the equation and fits in the form of $y=a{t}^2+bt+c$ ,

  $h=-\left(\frac{1}{2}\right)\cdot g{t}^2+vt$

Where, $a=-\left(\frac{1}{2}\right)\cdot g$ , $b=v$ , $c=0$ .

We use the formula, $t=\frac{-b}{2a}$

  $\begin{array}{l}t=\frac{-b}{2a}\\ t=\frac{-v}{2\cdot \left(-\left(\frac{1}{2}\right)\cdot g\right)}\\ t=\frac{-v}{-g}\\ t=\frac{v}{g}\end{array}$

Substituting $t=\frac{v}{g}$ in the given equation,

  $\begin{array}{l}h=vt-\frac{1}{2}g{t}^2\\ h=v\left(\frac{v}{g}\right)-\frac{1}{2}g{\left(\frac{v}{g}\right)}^2\\ h=\frac{v^2}{g}-\frac{v^2}{2g}\\ h=\frac{2v^2}{2g}-\frac{v^2}{2g}\\ h=\frac{v^2}{2g}\end{array}$

Hence, the greatest height of the projectile is $\frac{v^2}{2g}$ proved.