Velocity Question & Acceleration Below I have a question that I tried to solve on a exam. I am curious as to the actual way to approach the question. What I did was set the equation equal to $0$ and get $t = -3$ then I plugged in $3$ into the derivative of the equation and I got $128$ ft/sec. But apparently I approached it wrong. What is the right way to approach the question and solve it for the questions below?

If a rock is thrown upward on Earth, with an initial velocity of 32
  ft/sec from the  top of a 48 ft building, we can model the height of
  the rock at time t  by     $s(t) = −16t^2+ 32t + 48$. After throwing
  the rock up, it will at some point be  level again with the top of the
  48 ft building.  
What will the velocity be at this time?  
Find the acceleration, why is it constant?

 A: 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.
A: Note that we actually don't need to know $t$. By symmetry, the velocity when the rock reaches the "throw" level again must be the negative of the throw velocity. 
Since the throw velocity is $32$, the answer to the first question must be $-32$ (feet per second). 
A: When the rock is level again with the building, $s(t) = 48$.  So solving for time,
$-16t^2 + 32t = 0$ gives $t=0,2$.  So it appears that you just made a mistake in your calculation for $t$.  Your method for finishing the question seems correct, just using $t=2$ instead.
