A ball is thrown into the air with an initial upward velocity of 42 ft/s. It height h in feet after t seconds is given by the function h (t)=-16t^2+42t+4. What is the ball's maximum height?
We have the form
h(t) = at^2 + bt + c
The time for the max height is -b/ ( 2a) = = -42/ ( 2 * -16) = -42 / -32 = 21/16 sec
And the max height is - 16 ( 21/16)^2 + 42(21/16) + 4 = 31.5625 ft
