maximum ball height and time for it to reach it 

The given equation of displacement of the ball is 
$$s(t)=-16t^2+64t+8$$
The maximum height is obtained when velocity of the ball is zero; and velocity expression is given by
$$v(t)=s'(t)=-32t+64$$
So, equation velocity equals to $0$ we get the time which it takes to reach maximum height.

a) $$-32t+64=0$$ $$t=\boxed{2\;sec}$$

b) The maximum height will be given when we put time $t=2$ in the displacement expression,
$$s_{max}=s(2)=-16(2)^2+64(2)+8=\boxed{72\;ft}$$