Question

A ball is thrown up in the air, and its height as a function of time can be written as h(t) = -16t2 + 40t + 8.At what time is the ball at it's maximum height? Round to the nearest hundredth if needed.

Answer 1

hello

to solve this question, we simply need to solve for t

at maximum height,

$\begin{gathered} 0=-16t^2+40t+8 \\ 16t^2-40t-8=0 \\ \text{solve for t} \\ a=16,b=-40,c=-8 \\ t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ t=\frac{-(-40)\pm\sqrt[]{(-40)^2-4\times16\times-8}}{2\times16} \\ t=\frac{40\pm\sqrt[]{1600+512}}{32} \\ t=\frac{40\pm\sqrt[]{2112}}{32} \\ t=\frac{40\pm45.96}{32} \\ t=\frac{40+45.96}{32}=2.7s \\ or \\ t=\frac{40-45.96}{32}=-0.186s \end{gathered}$

since we can't have a negative value for time, the time to reach maxium heigh is equal to 2.7 seconds