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.
First you find the height after 2 seconds, we'll denote that as $y_2$. So using the fomula we have:
$$y_2 = 40 \cdot 2 - 16 \cdot 2^2 = 80 - 64 = 16$$
After 0.5 seconds, the time passed after the throw will be 2.5 seconds so for this height we have:
$y_{2.5} = 40\frac 52 - 16 \left(\frac 52\right)^2 =100 - 100 = 0$$
So after 2.5 second the height of the ball will be $0\text{ ft}$.
We know that the average velocity can be calulated using the following formula:
$$v = \frac{\Delta s}{\Delta t} = \frac{|y_1 - y_0|}{|t_1 - t_0|}$$
So after the substitution we have:
$$v = \frac{|0 - 16|}{|2.5 - 2|} = \frac{|-16|}{|0.5|} = \frac{16}{0.5} = 32$$
So the average speed in that period will be 32 $\frac{ft}{s}$
Now you can do the 0.1 second interval by yourself.
Best Answer
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 $-10$.