How do you solve this Calculus I problem

I don't need the answer,,, I need the steps one would take to solve this problem. Broken down to their smallest possible steps.

If a ball is thrown into the air from the top of a building 650 feet above the ground with an initial velocity of 32 feet per second, then its height in feet after t seconds is s(t) = 650 +32t - 16t^2. What will be the velocity of the ball when it hits the ground? What will be the acceleration of the ball when it hits the ground?

First of all, let's find the time that it takes for the ball to hit the ground, that is, when s = 0:
s(t) = 650 + 32t - 16t^2 = 0
t = (-32+-sqrt(32^2-4*650*-16))/(2*-16)
t = -5.45 or 7.45
Thus, t = 7.45s as t must be positive.

Now we need to find an equation for the velocity. The velocity is the rate of change of the position, relative to time; thus, it is the derivative of s(t):
v(t) = s'(t) = 32 - 2*16t = 32 - 32t
The velocity at t = 7.45 is:
v(7.45) = 32 - 32*7.45 = -206.46 ft/s or 206.46 ft/s downwards.

The acceleration is the rate of change of velocity:
a(t) = v'(t) = -32 ft/s^2

When the ball touches ground velocity becomes zero since it comes to a halt
ie. ds/dt=v=0 differentiating and putting ds/dt=0 we get t=r sec and differentiating again we get
d^2s/dt^2=a=-32 ft/sec^2 since acc. Is negative therefore s is max when t=1 sec. Substituting this value in eq of s we get max S=650+16=666ft.