The Physics Philes, lesson 4: Free Falling

In which more math is done, coins are dropped, and a vacation is taken.

Has it really been four weeks since I started this foolhardy journey into physics-land? My goodness, how time flies! I think I might be getting the hang of it. You’ll let me know if I’m messing up, right? Good. Because today I am again doing some mathy things: constant acceleration and free falling objects!

A couple of weeks ago I tried my best to explain acceleration and velocity. Today we have yet another permutation of that concept. Constant acceleration is when the velocity of an object changes at the same rate throughout the motion. Falling bodies have a constant acceleration if the effects of air are unimportant. (Note: For the purposes of today’s post I’ll consider air friction unimportant. It’s so much easier that way.)

OK, so I know what you’re thinking. What does this mean in math. Well I don’t really like your attitude, but I’ll try to explain it anyway. FOR SCIENCE.

If the x-acceleration is constant, that means the average acceleration for any time interval is the same as ax. Or…

ax = v2x – v1x / t2 – t1

That should look familiar. It’s average acceleration. But if acceleration is constant, it should be the same everywhere along the line. Get it? OK. But we’re not done. If t1 = 0 and t2 = any later time, then…

ax = vx – v0x / t2 – 0

vx = v0x + axt

So, x-acceleration is the constant rate of change of the x-velocity. The product of the change in the x-velocity per unit of time is axt and that equals the total change in x-velocity from the initial time to the later time. The x-velocity at any time equals the initial x-velocity plus the change in x-velocity.

Does any of that make even a little sense? Well I’m going to move on anyway. Because we need to be able to figure out where an object is!

We start with the equation for average velocity…

vav-x = x – x0 / t

The average velocity is the difference between the later position from the original position divided by time. The average x-velocity during any time interval is the average of the x-velocities at the beginning and the end of the interval…

vav-x – v0x + vx / 2

(This is only valid when the x-acceleration is constant.)

With a constant x-acceleration, the x-velocity at any time is…

vx = v0x + axt

If we plug that equation into the second vav-x equation, we get…

vav-x = 1/2(v0x + v0x + axt = v0x + 1/2axt

Oh man! But now we have two average velocity equations! we can set those equal to each other and get…

v0x + 1/2axt = x – x0 / t


x = x0 + v0xt + 1/2axt2

Or, in real people speak, if at any time t=0 a particle is at position x0 and has an x-velocity of v0x, it’s new position x at any later time is the sum of the initial position, the distance that it would move if x-velocity were consistent, and the additional distance caused by the change in velocity.

That’s all well and good, but how does it work? let’s do a quick example.

This example involved a free falling body. A free falling body is an object that is influenced by the force of gravity alone. Gravity is represented by g. Acceleration due to gravity is 9.8 m/s2. Gravity is the magnitude of a vector quantity, so it’s always positive.

Let’s drop a coin from the top of the Leaning Tower of Pisa. It starts from rest and falls freely. What is its position and velocity at 1.0 s?

First, let’s draw a diagram.

Notice that, since we’re talking about a vertical motion I used the y-axis. That means that I’ll replace the x’s in the equations with y. Don’t freak out on me. I have written down the known quantities and the target variables. We know that at t=0, y=0 and v0=0. Our target variables are v1y and y at 1 s. We’ll use the equations discussed today to figure this out.

Let’s first find the velocity of the coin at 1 s. The equation we need to use is…

vy = v0y + ayt

We know the initial velocity of the coin is 0 m/s, we know the time we are looking at is t=1 s, and we know that acceleration is -9.8 m/s2. (It’s negative because the coin is falling down in the negative y direction.) The equation now looks like this…

vy = 0 + (-9.8 m/s2)(1.0 s)

vy = -9.8 m/s

Now let’s find the position of the coin at 1 s. The equation we need to use is…

y = y0 + v0xt + 1/2axt2

Again, we just replace the letters with the numbers. We know both the initial y position and y-velocity are 0, we know the acceleration is -9.8 m/s2, and we know that t=1 s. We just gotta plug’em in and solve for y.

y = 0 + 0 + 1/2(-9.8 m/s2)(1.0 s)2

y = -4.9 m

Look at what we did! We calculated that a coin dropped from the top of the Leaning Tower of Pisa will fall 4.9 m at a velocity of 9.8 m/s after 1.0 s. That’s pretty cool!

As ever, I need to implore you for further clarification and/or correction. I’m sorry I haven’t been able to engage very much with commenters on this project, but please know that your thorough knowledge of this subject is impressive and appreciated.

Also, a programming note. Since I’ll be at SkepchickCON this weekend it’s likely I’ll be too tired to write a Physics Philes post. But I’ll keep on studying and, if you’re good, I’ll be back in two weeks with a little Isaac Newton in tow.

Featured image credit: We Know Memes

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Mindy is an attorney and Managing Editor of Teen Skepchick. She hates the law and loves stars. You can follow her on Twitter and on Google+.


  1. July 2, 2012 at 12:19 pm —

    I haven’t looked around enough to see if this suggestion has already been made, but you might want to consider using LaTeX, or another program like it. It really makes your equations a lot easier to read (it will look like what you would write down on paper), especially once you start using more complicated equations. And it’s FREE! You’ll have to master the commands, but it’s worth it in the end.

  2. July 2, 2012 at 6:04 pm —

    The picture used above–is it from one of the movies, or is it just a photoshop? I don’t recall Tony Stark every getting in a sports car in _The Avengers_, and Ruffalo wasn’t in any of the earlier movies, was he?

    • July 2, 2012 at 7:42 pm —

      Ruffalo wasn’t in any of the other movies. I want to say that it might be just a photo from on set, but not in the movie. Didn’t Bruce get into a convertible with Tony at the end of The Avengers? I’m not sure. What I am sure about is how much I love that picture 🙂

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