The given equation is:
[tex]y=-16t^2+40t+2[/tex]
It is required to find which interval or intervals of time will the projectile be above 18 feet.
To do this, solve the inequality:
[tex]\begin{gathered} y>18 \\ \Rightarrow-16t^2+40t+2>18 \end{gathered}[/tex]
First, find the critical points of the inequality by solving the equation:
[tex]\begin{gathered} −16t^2+40t+2=18 \\ \text{ Subtract 18 from both sides:} \\ \Rightarrow-16t^2+40t+2-18=18-18 \\ \Rightarrow−16t^2+40t-16=0 \\ \text{ Factor the left-hand side of the equation:} \\ \Rightarrow−8\left(2t−1\right)\left(t−2\right)=0 \\ \text{ Equate the factors to 0 to find the t-values:} \\ \Rightarrow(2t-1)=0\text{ or }(t-2)=0 \\ \Rightarrow2t=1\text{ or }t=2 \\ \Rightarrow t=\frac{1}{2}=0.5\text{ or }t=2 \end{gathered}[/tex]
The possible interval of solutions are:
[tex]t<0.5,\;0.52[/tex]
Use test values in the intervals to check which interval whose set of values satisfies the given inequality.
The only interval that satisfies it is 0.5.
Hence, the answer is between 0.5 second and 2 seconds.
The answer is option (c).
