Answer :
a) 1 (option 1)
b) 0.71 feet (option C)
Explanation:a) Given model:
[tex]h=-16t^2+10t\text{ + 3}[/tex]When height = 2 feet
We insert the value of the height in order to determine the constant in the equation:
[tex]\begin{gathered} 2=-16t^2+10t\text{ + 3} \\ \text{collect like terms by subtracting 2 from both sides:} \\ 0\text{ = }-16t^2+10t\text{ + 3}-2 \\ -16t^2+10t\text{ + }1\text{ = 0} \end{gathered}[/tex]The constant in the equation above is 1.
Hence, the constant in the quadrtic function is 1 (option 1)
b) when h = 2 feet, t =?
[tex]\begin{gathered} 2=-16t^2+10t\text{ + 3} \\ 16t^2-10t\text{ - 3}+2\text{ = 0} \\ 16t^2-10t\text{ - 1= 0} \\ u\sin g\text{ almighty formula as it can't be factorised} \\ a\text{ = 16, b = -10, c = -1} \end{gathered}[/tex][tex]\begin{gathered} t\text{ = }\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ t\text{ = }\frac{-(-10)\pm\sqrt[]{(-10)^2-4(16)(-1)}}{2(16)} \\ t\text{ = }\frac{10\pm\sqrt[]{100^{}+64}}{32} \end{gathered}[/tex][tex]\begin{gathered} t=\frac{10\pm\sqrt[]{164}}{32}=\frac{10\pm\sqrt[]{164}}{32} \\ t\text{ = }\frac{\text{10}\pm12.8}{32} \\ t\text{ = }\frac{\text{10+}12.8}{32}\text{or }\frac{\text{10-}12.8}{32} \\ t\text{ =}\frac{22.8}{32}\text{ or }\frac{-2.8}{32} \\ t\text{ = 0.7125 or -0.0875} \end{gathered}[/tex]We can't have a negative time, so t = 0.7125.
The time it takes to reach a height of 2 feet to the nearest hundredth is 0.71 feet (option C)