Finding the velocity of a rock given its height as a function of time I'm trying to learn calculus here, but I know I have to set the $h$ equal to 0 and find the time at when it's equal to 0, but I have no idea what to do after. Here is the question. How do I find out the velocity at that time?

If a rock is thrown upward on the planet Mars with a velocity of $10\;m/s$, its height (in meters) after $t$ seconds is given by $H = 10t − 1.86t^2$. Find the velocity of the rock when it hits the ground.

 A: Find the positive root of $H(t)=10t-1.86t^2$. Let $t_1$ be that root. Then evaluate $H'(t_1)$. 
Added: Plot of the height function $H(t)=10t-1.86t^2$ (in meters) vs. time $t$ (in seconds)

Added 2: Plot of $H(t)$ (black, in meters) and $H'(t)=10-3.72t$ (blue, in m/s) vs. time $t$ (in seconds)

A: 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 $-10$. 
A: These types of problems can all be solved by knowing the relationship between the position, velocity, and acceleration equations. In the following, by taking the derivative you can move from one equation to the next:
$$
\text{position} \to \text{velocity} \to \text{acceleration}
$$
Similarly, to go from one equation to the next below, you need to integrate the relevant equation:
$$
\text{acceleration}\to\text{velocity}\to\text{position}
$$
These equations make more sense when you know the general forms for each equation (assuming constant linear acceleration). The general position equation is given by
$$
s(t) = \frac{1}{2}at^2+v_0t+s_0
$$
where $a$ is acceleration, $t$ is time, $v_0$ is initially velocity, and $s_0$ is the initial position. The general equation for velocity is
$$
v(t) = at+v_0
$$
using the same variables. Finally, the equation for constant linear acceleration is simply
$$
a(t)=a
$$
Here you're given the height, or position, equation and you want to get the velocity. From what I wrote above, it's clear that you want to take the derivative to get the velocity. 
A: You have $H(t)=10t-1.86 t^2$ 
The rock hits the floor when $H(x)=0$, this is, when $0=10t-1.86 t^2$. 
It is clear the solution $t=0$ stands for the moment it is thrown, so we consider $0=10-1.86 t$, which leads to $t=\frac{1000}{186}$.
The speed of the rock is given by the magnitude of the velocity at the moment $t=\frac{1000}{186}$. We use the derivative, which is $H'(t)=10-3.72 t$. To get the value we evaluate $H'(\frac{1000}{186})=-10$
A: at h=0, the rock hits the ground.

$0=10t-1.86t^2$

using the quadratic formula to solve, t= 0s and t=5.37s - the time taken for rock to hit the ground.
$h'(x)=velocity$
$h'(x) =10-3.72t$,
and t=5.37 from above
and thus,
the velocity of the rock when it hits the ground is = 10-3.72(5.37)
thus, velocity = -9.98m/s
A: For $$H = 10t − 1.86t^2 = (10 − 1.86t)t$$ you have $$V = \frac {dH}{dt} = 10 − 2\cdot 1.86t = (10 − 1.86t) - 1.86t$$
Height is zero either when $t=0$ (the moment when the rock is thrown) or when it lands — so the landing time is when $(10 − 1.86t)$ is zero, that is at $$t=\frac {10}{1.86}$$
Then the velocity $$V_\text{landing} = (10 − 1.86t) - 1.86t = 0 - 1.86t = -10$$ opposite to $$V_\text{thrown} = 10 − 2\cdot 1.86\cdot 0 = 10$$
