The height ##h## (in feet) of a volleyball ##t## seconds after it is hit can be modeled by ##h = -16t^2+48t+4##. How long is the volleyball in the air?

##t = 3.08s##

We have to substitute ##h## by ##0##, as that is the height both when projected as when it hits the ground. Then we solve for ##t##: ##-16t^2 + 48t + 4 = 0##, ##16t^2 - 48t - 4 = 0##, ##4(4t^2 - 12t - 1) = 0##, ##4t^2 - 12t - 1 = 0##. Now we use ##t = (-b +- sqrt((b^2 - 4ac)))/(2a)##: ##t = (- (-12) +- sqrt((-12)^2 - 4(4)(-1)))/(2(4))##, ##t = (12 +- sqrt(144 + 16))/(8)##, ##t = (12 +- sqrt(160))/(8)##, ##t = 3.08s##.

Hope it Helps! :D .