Question

This is math homework, I don’t understand how to find the seconds after a balls hight

Answer 1

Substituting v₀=180 and h(t)=440 in the given equation we get:

$440=-16t^2+180t\text{.}$

Subtracting 440 from the above equation we get:

$\begin{gathered} 440-440=-16t^2+180t-440, \\ 0=-16t^2+180t-440. \end{gathered}$

Dividing the above equation by -2 we get:

$\begin{gathered} \frac{0}{-2}=\frac{-16t^2+180t-440}{-2}, \\ 0=8t^2-90t+220. \end{gathered}$

Using the quadratic formula for second-degree equations we get:

$\begin{gathered} t=\frac{-(-90_{})\pm\sqrt[]{(-90)^2-4(8)(220)}}{2\cdot8}, \\ t=\frac{90\pm\sqrt[]{8100-7040}}{16}, \\ t=\frac{90\pm\sqrt[]{1060}}{16}, \\ t=\frac{90\pm2\sqrt[]{265}}{16}, \\ t=\frac{45\pm\sqrt[]{265}}{8}\text{.} \end{gathered}$

Therefore, the firework will reach a height of 440 feer after

$\begin{gathered} t=\frac{45-\sqrt[]{265}}{8}\approx\frac{45-16.2788208}{8} \\ =\frac{28.7211794}{8}=3.590147425\approx3.6. \end{gathered}$

seconds.

Answer: 3.6.

The number without being rounded yet is 3.590147425.