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.
The equation:
$$ \frac{dv}{dt}+\frac{v}{40}=-g$$ is a first-order linear ODE with initial condition $v(0)=200$. The solution (via integration factor) is:
$$ v(t)=e^{-t/40}[200+40g]-40g.$$ At the maximum height the object's velocity will be zero (at the instant that it starts to come back down). Thus set $v=0$ and solve for $t$ to get $t=-40\ln{\frac{40g}{200+40g}}=16.49$ seconds. To determine at what height, $h$, this occurs, realize that $v=\frac{dh}{dt}$ and integrate the expression
$$ v=\frac{dh}{dt}=e^{-t/40}[200+40g]-40g$$ with the initial condition of $h(0)=0$ to get
$$ h(t)=40(200)+40^{2}g-40gt-e^{-t/40}[40(200)+40^{2}g]. $$ Now plug $t=16.49$ into this expression to find $h=1536$ meters.
Hope this helps.
Cheers,
Paul Safier
Best Answer
If you insist on deriving all from the formulas
\begin{align} v(t) &= v_0 + at \tag 1\\ s(t) &=s_0+v_0t+\frac{1}{2}at^2\tag 2 \end{align}
only, let you will be fulfilled.
(Note: $(2)$ is derived from $(1)$ by integration.)
First, the acceleration is negative, as the Earth attraction works against the movement, so $a = -g,\ $ and $(1)$ and $(2)$ become
\begin{align} v(t) &= v_0 - gt \tag 3\\ s(t) &=s_0+v_0t-\frac{1}{2}gt^2\tag 4 \end{align}
Now, to better understand the situation, substitute in $(3)$ known values $s_0 = 200\ \mathrm {ft}\,$ and $g \approx 32.17\ {\mathrm {ft}\cdot \mathrm s}^{-2},\ $ and (meanwhile) unknown value $v_0 \approx 113.4\ \mathrm{ft}\cdot {\mathrm s}^{-1}\ $ (which we will calculate at the end).
We get this graph - a linear dependency of a speed from time:
We may see that the speed of the projectile decreases from the speed $v_0 \approx 113.4\ \mathrm{ft}\cdot {\mathrm s}^{-1}\ \ $ (in time $t_0=0)\ $ to zero speed $v_1 = 0\ $ (in time $t_1 \approx 3.5\ \mathrm s)$.
Substituting the same values into $(4)$ we obtain this graph - a quadratic dependency of the position of the projectile from time:
We may see that the position of the projectile increases from the position $s_0 = 200\ \mathrm{ft}\ \ $ (in time $t_0=0)\ $ to the position $s_1 = 400\ \mathrm{ft}\ \ $ (in time $t_1 \approx 3.5\ \mathrm s)$.
So let $t_1$ is the time in which the projectile reaches its maximum height of $s_1 = s(t_1) \ $ (and we know that $s_1 = 400\ \mathrm{ft}.$) Equations $(3)$ and $(4)$ become
\begin{align} v(t_1) &= v_0 - gt_1 \tag 5\\ s(t_1) &= s_0+v_0t_1-\frac{1}{2}gt_1^2\tag 6 \end{align}
But $\ v(t_1) = 0\ $ and $\ s(t_1) = s_1$:
\begin{align} 0 &= v_0 - gt_1 \tag 7\\ s_1 &= s_0+v_0t_1-\frac{1}{2}gt_1^2\tag 8 \end{align}
From $(7)$ we obtain
$$v_0 = gt_1\tag 9$$
and substituting it into $(8)$ we obtain
$$s_1 =s_0+g{t_1}^2-\frac{1}{2}gt_1^2$$
i. e.
$$s_1 =s_0+\frac{1}{2}gt_1^2 \tag {10}$$
Now it is all known, except $t_1$, so let express it from $(10)$:
$$t_1 = \sqrt{{2(s_1-s_0)\over g}}$$
Substituting it into $(9)$ gives us the result
$$v_0 = gt_1 = g\sqrt{{2(s_1-s_0)\over g}} = \sqrt{{2(s_1-s_0) g}}$$
in particular
$$v_0 = \sqrt{2\times 200 \times 32,17}\ {\mathrm {ft}\over\mathrm m} \approx 113.4 {\text{ft}\over\text{m}}$$