To determine the equation of the ball’s height, repeat the procedure in part (a), but integrate the
velocity function to get the position function: v(t) = = −64 + 160. So, ∫ dh = ∫ (−64t + 160)
dt or h(t) = −32t^2 + 160t C. To find C, plug in the ball’s height at time t = 0, 100 ft, and C =- Thus, h(t) = −32t^2 + 160t + 100.
(d) The ball’s maximum height?
The maximum height occurs when the velocity of the ball is zero—when it is changing
direction from rising to falling. In part (b), we found that time to be t = 2.5 sec. Plug 2.5 into
the position function to solve for the maximum height: h(2.5) = −32(2.5)^2 + 160(2.5) + 100 =
300ft.