Given:
[tex]y=-16x^2_{}+170x+61[/tex]
Where x represents the time in seconds and y represents the height.
To find the time that the rocket will hit the ground:
Put y=0,
[tex]-16x^2_{}+170x+61=0[/tex]
Using the quadratic formula,
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ =\frac{-170\pm\sqrt[]{(170)^2-4(-16)(61)}}{2(-16)} \\ =\frac{-170\pm\sqrt[]{28900+3904}}{-32} \\ =\frac{-170\pm\sqrt[]{32804}}{-32} \\ =\frac{-170\pm181.118}{-32} \\ x=\frac{-170+181.118}{-32},x=\frac{-170-181.118}{-32} \\ x=-0.3474,10.9724 \\ x\approx-0.347,10.972 \end{gathered}[/tex]
Hence, the time required to hit the ground in 10.972 seconds.
