First we find the time(s) when the rock is at ground level. So set $10t-1.86t^2=0$ and solve for $t$. We get $t=0$ and $t=\frac{10}{1.86}$.
The velocity at time $t$ is the derivative of the displacement function $H(t)$. So the velocity at time $t$ is $10-(2)(1.86)t$. Substitute the value of $t$ we found above.
Remark: We can solve the problem instantly without calculus. The initial velocity is $10$. So by symmetry the velocity when it hits the ground on its return trip must be $-10$.
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.
Best Answer
The question asks us to find the velocity. The velocity at any time $t$ is equal to $\frac{dh}{dt}$. We have in general $$\frac{dh}{dt}=15-3.72t.\qquad\text{(Equation 1)}$$
If we had been asked for the velocity at time $t=3$, for example, life would be easy, we would just substitute $3$ for $t$ in the above equation.
Unfortunately, we have been asked something more complicated, namely the two possible velocities when the height of the rock is $25$.
The height $h$ of the rock is $25$ when $$25=15t-1.86t^2. \qquad\text{(Equation 2)}$$
From this equation, we should be able to find the two times $t$ when $h=25$. And once we know these two times $t$, simple substitution in Equation 1 will give us the velocities.
Equation 2 is equivalent to $1.86t^2-15t+25=0$. This is a standard quadratic equation, with slightly messy coefficients. To solve for $t$, we use the Quadratic Formula. We get $$t=\frac{15\pm\sqrt{15^2-(4)(1.86)(25)}}{3.72}.\qquad\text{(Equation 3)}$$
Finally, we use the calculator to find good approximations to the two times $t$ when the height of the rock is $25$, and substitute in Equation 1.
Comment: We suggested using the calculator to find the two values of $t$ at which the height is $25$. But there is a better way. The velocity at time $t$ is $15+3.72t$. Substitute the values of $t$ obtained in Equation 3. There is very nice simplification, and we get that the two velocities are $\pm\sqrt{15^2-(4)(1.86)(25)}=\pm\sqrt{39}$. Thus it turns out (by symmetry, this is no big surprise!) that the speed of the rock is the same when it reaches height $25$ on its way up as when it reaches height $25$ again on its way down. But the velocities (positive for up, negative for down) are different. The great simplification that we got by not using the calculator immediately hints that there may be a simpler approach to the problem. And indeed there is, but the approach that we described in the main part of the answer is the most straightforward one. However, it can be in general quite useful not to bring out the calculator too early.