Question

A firework rocket is shot upward at a rate of 640 ft/sec. Use the projectile formula $\mathrm{h}=-16{\mathrm{t}}^2+{\mathrm{v}}_{\mathrm{o}}\mathrm{t}$to determine when the height of the firework rocket will be 1200 feet.

Short answer

Time will be 38 second and 2 second.

Step-by-step solution

Step 1. Given Information

The initial speed is 640 feet/sec. The height of the rocket is 1200 feet.

Step 2. Solving equation

On applying the equation,

$\mathrm{h}=-16{\mathrm{t}}^2+{\mathrm{v}}_{\mathrm{o}}\mathrm{t}$

On simplifying,

$$\begin{gathered} -16{\mathrm{t}}^2+640\mathrm{t}=1200 \\ -16{\mathrm{t}}^2+640\mathrm{t}-1200=0 \\ {\mathrm{t}}^2-40\mathrm{t}+75=0 \\ \mathrm{a}=1,\mathrm{b}=-40,\mathrm{c}=75 \\ \mathrm{t}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}} \\ \mathrm{t}=\frac{-(-40)\pm \sqrt{{(-40)}^2-4\times 1\times 75}}{2\times 1} \\ \mathrm{t}=\frac{40\pm \sqrt{1600-300}}{2} \\ \mathrm{t}=\frac{40\pm 10\sqrt{13}}{2} \\ \mathrm{t}=20+5\sqrt{13}=20+18.02\approx 38 \\ \mathrm{t}=20-5\sqrt{13}=20-18.02=1.98\approx 2 \end{gathered}$$

The time is 38 second and 2 second respectively.