Recall that for constant acceleration only, the position of the object is given by:
$s = s_0 + v_ot + \frac {1}{2} t^2 \space \text{where v = velocity, s = position} \tag{1}$
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, $s''(t)$ or rather $a$ for acceleration 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.