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! 😀 .