Question

I need help with this question. Do I equal h(t) to 10 and then solve?

Answer 1

incSolution:

Given that;

$h(t)=-16t^2-10t+50$

To fall 10 feet off of the ground,

Where

$h(t)=10\text{ ft}$

Substitute for h(t)

$\begin{gathered} 10=-16t^2-10t+50 \\ -16t^2-10t+50=10 \\ -16t^2-10t+50-10=10-10 \\ -16t^2-10t+40=0 \end{gathered}$

Applying quadratic formula

$\begin{gathered} t=\frac{-\left(-10\right)\pm\sqrt{\left(-10\right)^2-4\left(-16\right)\cdot\:40}}{2\left(-16\right)} \\ t=-\frac{5+\sqrt{665}}{16},\:t=\frac{\sqrt{665}-5}{16} \\ t=-1.92422\text{ or }t=1.29922 \end{gathered}$

Since, t can not be negative,

The time it will take is 1.30 seconds (nearest hundredth)