# 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.

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What are the steps i take primarily to find it after? Or if anyone could show me how its done hat would be awesome. –  soniccool Jun 17 '12 at 20:41
Can you find velocity as a function of $t$? –  user17794 Jun 17 '12 at 20:44

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 $-10$.

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Alright i tried it out myself, it works. So after i solve it i dervice the original equation and just plugin the values! –  soniccool Jun 17 '12 at 21:00

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.

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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)

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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 $H'(\frac{1000}{186})=-10$

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The $t$ you're looking for is actually $\frac{1000}{186}$ which gives the correct answer of $-10$ m/s. –  chris Jun 17 '12 at 21:00
@chris Sorry, I'll correct that, my fault. –  Pedro Tamaroff Jun 17 '12 at 21:10