Question

How long after the arrow was shot does it hit its intended target

Answer 1

`\begin{gathered} h(t)=-16t^2+100t+5 \\ at\text{ }20ft\text{ above the ground h =20ft} \\ \Rightarrow-16t^2+100t+5=20 \\ \Rightarrow16t^2-100t+15=0 \\ \Rightarrow t^2-(25)/(4)t+(15)/(16)=0 \\ \Rightarrow(t-(25)/(8))^2-((25)/(8))^2+(15)/(16)=0 \\ \Rightarrow(t-(25)/(8))^2=(625)/(64)-(15)/(16)=(565)/(64) \\ \Rightarrow t=(25)/(8)\pm\sqrt[]{(565)/(64)} \\ t=\frac{25\pm\sqrt[]{565}}{8} \\ t=0.15s\text{ or 6.10s} \end{gathered}`

The smaller time 0.15s is the time taken for the arrow to reach 20ft when it was moving upward while the bigger time 6.10s is the time taken for the arrow to reach 20ft when it was coming back down.

So it took the arrow 6.10s to hit its intended target