ball is thrown upwards at 20 ft/sec, the height h after t seconds is given by h = -16t^2 + 20t. How many seconds does it take for the ball to reach its maximum height, and what is this height?

Answer 1

The vertex of this downwards-opening parabola lies halfway between the t-intercepts
these can be found quite easily:

h = -4t(4t - 5)

so t = 0 or t = 5/4
so halfway would be t = 5/8

when t=5/8, h = -16(25/64) + 20(5/8) = 6.25

so the max height of 6.25 ft will be reached after 5/8 seconds