In this problem the function that models the hight is:
[tex]h(t)=-5t^2+14t+3[/tex]So when the water hits the grownd h=0 so we replace that and solve for t so:
[tex]0=-5t^2+4t+3[/tex]To solve this expression we can use the cuadratic equation:
[tex]x=\frac{-4\pm\sqrt[]{16-4(-5)(3)}}{-10}[/tex]and we operate so:
[tex]\begin{gathered} x=\frac{-4\pm\sqrt[]{76}}{-10} \\ x=\frac{-4\pm8.7}{-10} \end{gathered}[/tex]Now we solve bout of the equation so:
[tex]\begin{gathered} x_1=\frac{-4+8.7}{-10}=-0.47 \\ x_2=\frac{-4-8.7}{-10}=1.27_{} \end{gathered}[/tex]So the answer that have sense is the secon one so the water hits the ground after 1.27 seconds
