Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 51

At $t=2$.

The stone will reach its maximum height when its velocity is zero, which occurs at the only critical point for this inverted parabola. We have that $v(t) = s(t) = −32t + 64$, which is zero when $t = 2$. The height at this time is $s(2) = 256$, the maximum height.