Velocity word problem I know how to find velocity but I just can't make sense of this problem. 
If a rock is thrown verically upward from the surface of mars with velocity 15m/s, its height after t seconds is $h=15t - 1.86t^2$
What is the velocity of the rock when its height is 25 m on its way up and its way down.
So when I think I need to do is set it equal to 25 so $25=15t-1.86t^2$ and then from that I should get a positive and a negative answer maybe? I am not sure what to do from here. I know I will need to find the derivative but not sure how.
 A: 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.
A: You presumably know some formulae which look something like $s = ut +\frac{1}{2}at^2$ and $v=u+at$, where $s$ is distance at time $t$, $u$ is the initial velocity, $v$ is the velocity at time $t$, and $a$ is the constant acceleration.  
So if $s=h$, you can find what the initial velocity $u$ and acceleration $a$ are. 
You could solve $25 = 15t - 1.86 t^2$ for $t$ to get two solutions, and then use them to find the two values of $v$.  
A: You have something of the right start with plugging in 25 for h, but you've taken the derivative too early.  Remember that the position (height) equation is given in terms of t. Thus its derivative, whose output will give you velocity, is also in terms of t.  You found the correct derivative $\frac{dh}{dt} = 15-2(1.86t)$.  So, all you need now, is the values of t such that h is 25.  
HINT: t will be the solutions of the equation you gave above: $25=15t-1.86t^2$
Can you go from there?
