A ball is thrown into the air with an initial upward velocity of 48 ft/s. It height h in feet after t seconds is given by the function h (t) = – 16t2 + 48t + 4.
A) What height will the ball be when 2 seconds has passed?
t = time Just put t = 2 in to the equation and calculate h
h = - 16(2^2) + 48(2) + 4 Got it ?
Thank you for the help, I don't know why I missed something as simple as that.