Physics on Mars

I have always been awestruck by physics. I wanted to understand how the equations worked, and how we could relate math to the universe. It is indeed spectacular. In this post, I will walk through a calculus problem on Mars. It is quite fun to see how calculus can help us solve real life problems. Without further ado, let us jump right in…

Alright, this is so exciting. What I enjoy doing before solving any physics problem is drawing a picture of the problem to get a better understanding. I am allowed to let my long-lost childhood artistic talent on the loose:

Now that we have a nice picture, we can go ahead and start solving the following:

a) Find the velocity function v(t).

Well, we know the gravitational force, and gravity is nothing but acceleration. The rock feels the acceleration due to gravity, so we have a(t)= -12. This is great because velocity is the integration of acceleration:

We have found v(t) = -12t + c1. But what is c1? We can find c1 (which is a constant found in the process of integration) by replacing v(t) by its value of 96 ft/s and letting t=0, since 96 ft/s is the initial velocity, meaning that the time is zero:

Therefore, the velocity function is v(t)= -12t+96. Now onto the next part…

b) Find the position function s(t).

Just like the previous part, we can find the position function by solving the integral of the velocity function:

Moreover, we must find c2 at the initial position (when time is zero):

Thus, the position function is s(t) = -6t^2+96t. And lastly…

c) Determine when the rock hits the ground.

Now we get to make sense of these equations. To determine when the rock hits the ground, we may use the position function. Intuitively, we have two variables: y and t. These are position (height) and time respectively. When the rock is on the ground, its position is zero. The independent variable — time — is what can change. Think about this; the initial position of the rock is zero since it is at height zero (y=0). Then after being launched, the rock will eventually hit the ground since Mars’s gravity is pulling it down, then it will be at position zero again. Now let us see this in an equation:

First we let the position be zero and solve for the time…

Then we get two answers:

This proves our theory that the rock will be on the ground right before launching, and after hitting the ground post launch. But we already knew that t=0 was the initial time, so the answer to c) is t=16 or 16 seconds. Of-course, we can always check our work to make sure it is correct. First let us use our answer directly:

So after 16 seconds, the rock hits the ground, perfect. We can also go to Desmos and see our rock flying from start to finish. All we have to do is type in our position function and look at the values on the graph:

This graph is the trajectory that the rock on mars will take in this scenario. This problem was super pleasant to showcase, thank you for reading.