Solving the quadratic equation, it is found that the diver was in the air for 2 seconds.
The height of the diver after t seconds is given by:
[tex]h(t) = -16t^2 + 8t + 80[/tex]
He is in the air until [tex]h(t) = 0[/tex], hence:
[tex]-16t^2 + 8t + 80 = 0[/tex]
[tex]16t^2 - 8t - 80 = 0[/tex]
Which is a quadratic equation with coefficients [tex]a = 16, b = -8, c = -80[/tex]. Then:
[tex]\Delta = (-8)^{2} - 4(-16)(80) = 5184[/tex]
[tex]x_{1} = \frac{-8 + \sqrt{5184}}{32} = 2[/tex]
[tex]x_{2} = \frac{-8 - \sqrt{5184}}{32} = -2.5[/tex]
Time is a positive measure, hence we take the positive solution, which means that the diver was in the air for 2 seconds.
A similar problem involving the solution of a quadratic equation is given at https://brainly.com/question/24764843
