A stone thrown vertically upward satisfies the equation $$s = 64t - 16{t^2},$$    where $$s$$ is in meter and $$t$$ is in second. What is the time required to reach the maximum height?

Solution :
Given equation is $$s = 64t - 16{t^2}$$
$$\therefore $$  On differentiating w.r.t. $$t,$$ we get $$\frac{{ds}}{{dt}} = 64 - 32t$$
Put $$\frac{{ds}}{{dt}} = 0$$   for maximum height
$$\eqalign{ & \Rightarrow 64 - 32t = 0 \cr & \Rightarrow t = 2 \cr} $$
Now, $$\frac{{{d^2}s}}{{d{t^2}}} = - 32$$
At $$t = 2,\,\frac{{{d^2}s}}{{d{t^2}}} = - 32$$
Since, $${\left( {\frac{{{d^2}s}}{{d{t^2}}}} \right)_{t = 2}} < 0$$
$$\therefore $$  Required time $$= 2$$  seconds