Recall that for constant vertical acceleration only, the position of the object is given by:
$$s\left(t\right) = s_0 + v_ot + \frac {1}{2} a t^2,$$
with constant vertical acceleration $a$, initial velocity and position $v_0$ and $s_0$, respectively, and current vertical position $s$.
Since our initial height and final height are both $48$ feet, we have:
$$48 = 48 + 32t - 16t^2 \tag{2}$$
Solving for the time to where the rock is at the same height it started at leads us to:
$$0= -16t^2 + 32t \implies t = 0, 2 \space \text{sec} \tag{3} $$
Obviously the rock is at the same height at $t = 0$, so at $t = 2$ sec is where it makes a parabola-like movement and comes back down to the original height of the $48$ foot building.
Taking the derivative of our position function, $s(t)$ and plugging in $t = 2$ sec leaves us:
$$s'(t) = -32t + 32 \implies v = -32 \space \text{feet/sec} \tag{4} $$
Taking the derivative again leads us to:
$$s''(t) = -32 \space \text{feet per second per second} \tag{5}$$
As expected, $s''(t)=a$ is indeed constant, since the ball is only accelerating due to gravity which is roughly $-32$ feet per second per second. Also, you could note that since acceleration is the second derivative of a position function and our position function was a quadratic, it follows that the acceleration function is constant since taking the derivative each time lowers the power. Further, as noted in equation (1), the formula only holds for constant acceleration, so it is a bit circular to see that since your given equation fits the model formula (1), the object undergoes constant acceleration.
You’re using the word acceleration to mean two different, though related, things.
In colloquial usage (what you all “real life”), to accelerate means to increase speed, and to decelerate means to decrease speed. However, you’re working with velocities, which are directed—here signed—quantities. In this context, the word “acceleration” refers to any change in velocity, regardless of whether it increases or decreases the object’s speed, which is the velocity’s absolute value. This acceleration is also a signed quantity, but to determine whether it corresponds to an increase or decrease in speed, you also have to consider the sign of the velocity: when the signs are the same, the speed increases—an “acceleration” in the first sense above; when the signs differ, the speed decreases—a “deceleration.”
Best Answer
The acceleration you get after differentiation is actually the instantaneous acceleration, and so you cannot equate the average acceleration directly.
However, if you do the average of the instantaneous accelerations at the two end points, that'd work. Here, it will be $\dfrac12(2+0)=1$ as you expected.
There is no constant missing, the definition is correct. And this average part works only in certain cases, for instance when your original function is a quadratic, meaning the derivative is linear. Since the derivative is linear, it increases with a constant rate, and so you can just take the average rate between the end points.
Consider a linear sequence or an arithmetic progression, $a_n=bn+c$, then the average of the first $n$ terms is $$\dfrac{S_n}{n}=\dfrac12(a+a_n)$$ which is just the average of the first and last terms.
Hope this helps. Ask anything if not clear. :)