If you know the maximum height, the answer is really simple to find - we can directly work from there, knowing that the velocity at maximum height is $0 \frac m s$ because it is, essentially, not moving as it turns from upwards motion to downwards free-fall.
Note that there are two approaches to finding this solution - one, more basic and perhaps easier to understand (but longer and more complex), the kinematic approach, using basic ideas of velocity, acceleration, and displacement - and the other, although more advanced and, much simpler - the energy approach. I'll start with the basic solution using the kinematic approach.
I. The Kinematic Approach
We know that, for any object under constant acceleration, the velocity at any point in time can be modeled by
$$v = v_0 + at$$
As previously stated, because the ball falls starting from its maximum height, the initial velocity is $0$. Therefore, the equation becomes
$$v = at$$
where $v_0$ is the initial velocity, $a$ is acceleration, and $t$ is the time elapsed. For most free fall cases, $a = -g$ (where $g$ is gravitational acceleration) as gravitational acceleration is downwards. However, in this particular case, we already know the maximum height and only need to examine the part of the trajectory that is downwards - therefore, if we simply "rotate our axes," setting the direction of motion as the positive direction, we can set $a = g$ because gravity would be working in the same direction of our current motion. Therefore, we can modify the equation to become
$$v = gt$$
Now all that's left is finding the time elapsed to reach the ground. We know, first of all, that we can find the displacement of an object under constant acceleration by averaging the initial and final velocities and multiplying that by time, thus
$$\Delta x = \frac {t(v_0 + v)} 2$$
Again, here, we know $v_0 = 0$, so the equation becomes
$$\Delta x = \frac {t(v)} 2$$
From here, we can solve for $t$ and then substitute our solution for $t$ into the previous equation to solve for $v$.
\begin{align*}
\Delta x &= \frac {t(v)} 2 \quad && \\
\frac {2\Delta x} v &= t \quad && \text{Multiply both sides by $\frac 2 v$} \\
t &= \frac {2\Delta x} v \quad && \text{Solve for $t$}
\end{align*}
Then, we substitute that into the previous equation:
\begin{align*}
v &= gt \quad && \text{Previous equation} \\
v &= g(\frac {2\Delta x} v) \quad && \text{Substitute for $t$} \\
v &= \frac {2g\Delta x} v \quad && \text{Simplify fractions} \\
v^2 &= 2g\Delta x \quad && \text{Multiply both sides by $v$} \\
v &= \sqrt{2g\Delta x} \quad && \text{Square root both sides}
\end{align*}
Then, we can plug in everything we already know to solve for $v$. Because of personal preferences ;), I'll do it in SI units:
- $g = 9.81 \frac m {s^2}$ or $32.19 \frac {ft} {s^2}$, the gravitational acceleration
- $\Delta x = 132 ft$ or $40.23 m$, the maximum height (or total displacement)
\begin{align*}
v &= \sqrt{2 \cdot 9.81 \frac m {s^2} \cdot 40.23 m} \\
&\approx 28.04 \frac m s \\
&\approx 91.99 \frac {ft} s \text{ or } 92 \frac {ft} s
\end{align*}
II. The Energy Approach
Assuming a system consisting of just the ground (gravitational force) and the ball, we can apply the Law of Conservation of Energy, which states that, in any closed system, the net energy (kinetic + potential) is constant. Here, we need only to find the energy at the two endpoints of motion and equate the two.
Energy at Peak of Motion
As previously stated, the velocity is 0 at the peak of motion - therefore, the only energy present would be potential energy. We know that gravitational potential energy is given by $mgh$, where $m$ is the mass of the ball, $g$ is the gravitational acceleration, and $h$ is the height. Thus
$$E = mgh$$
Energy at Ground (Bottom)
Here, when the ball is at the ground, the height $h$ is essentially $0$ (except that tiny (negligible) bit, as it isn't quite in contact with the ground yet), and all that needs to be measured is kinetic energy. We know that kinetic energy is given by $\frac 1 2 mv^2$, therefore
$$E = \frac 1 2 mv^2$$
Conservation of Energy
Now, we can easily equate the energy representations at different points in the ball's trajectory at the two endpoints and solve for $v$.
\begin{align*}
mgh &= \frac 1 2 mv^2 \quad && \text{Equate the two energy representations} \\
gh &= \frac 1 2 v^2 \quad && \text{Divide both sides by $m$} \\
2gh &= v^2 \quad && \text{Multiply both sides by 2} \\
\sqrt{2gh} &= v \quad && \text{Square root both sides} \\
v &= \sqrt{2gh} \quad && \text{Solve for $v$}
\end{align*}
Unsurprisingly, this yields the exact same result as it would with the kinematics approach - but is a much shorter and more concise solution - the only downside to this solution is that it requires a deeper understanding of Physics concepts of work and energy.
Best Answer
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.