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.
Let $x$ denote the position along the chain in meters, with $x=0$ corresponding to the top of the chain at the top of the building. Let $\delta x$ denote a small length of chain. To figure out the total work, we break up the chain into small pieces of equal length $\delta x,$ compute the work to move each piece, and add it all together.
The top piece does not need to be lifted further. A piece at approximately $x$ meters down the chain needs to move $x$ meters if $x\le 3,$ or simply moves $3$ meters if $x\ge 3.$
The work required to lift a piece is $xF$ if $x\le 3,$ or $3F$ otherwise, where $F$ is the force due to gravity. Assuming the chain has constant density, the force acting on a small $\delta x$ of chain is $F=mg=(\frac{20}{64}\delta x)(9.8) = 3.0625 \delta x.$
So to move a small $\delta x$-length of chain takes $3.0625x\delta x$ Joules of work if $0\le x\le 3.$ Summing up the work over these small lengths corresponds to integrating $\int_0^3 3.0625xdx.$
The work required to move the rest of the chain is easier, since all the remaining $\delta x$-lengths move the same distance: $3$ meters. So we can compte this all at once as $3F=3mg=3(\frac{20}{64}61)(9.8)=560.4375$ Joules of work.
Best Answer
Notice, the maximum height $(h_{max})$ obtained by the rock above the building where the final vertical velocity becomes zero.
Hence, using third equation of motion (moving upwards) $$v_y^2=u_y^2-2gh_{max}$$ setting the corresponding values, we get $$0=(30)^2-2(32)h_{max}\implies h_{max}=\frac{225}{16}$$ Now, the rock starts falling down from the highest point at a height $h=h_{max}+200=\frac{225}{16}+200=\frac{3425}{16}$ from the ground & there the initial vertical velocity is zero i.e. $u_y=0$
Hence, using third equation of motion (falling downwards) $$v_y^2=u_y^2+2gh$$ setting the corresponding values, we get $$v_y^2=0+2(32)\left(\frac{3425}{16}\right)$$$$v_y=\sqrt{13700}$$ $$\color{red}{v_y=10\sqrt{137}\approx 117.047\ ft/s}$$