It’s raining Aquaman. Hallelujah?

The new Justice League trailer that debuted at the San Diego Comicon has a little bit of everything. Including a falling Aquaman.

Technically, he’s not falling, he’s standing on a parademon, and they’re both falling, but still. Details.

Things that fall all follow the same rules, and in terms of teaching physics, a falling Aquaman is a heck of a lot more interesting than a falling ball (or any other example that can be found in a textbook), so let’s use him as an example of a falling…*thing* and do some basic physics. Ride along with us, and I guarantee you’ll learn some physics.

While Aquaman is falling on his parademon surfboard, the combo are falling under the influence of gravity only. That’s called ** free fall**. No rockets, no motors, no parachutes – only gravity. All objects in free fall experience an acceleration downward of 9.8 meters/second per second. That means that for each second Aquaman is falling he’s adding 9.8 meters/second to his velocity. To indicate, mathematically, that Aquaman’s velocity is downward, we can use the value of -9.8 m/s/s or -9.8 m/s

^{2}as our acceleration.

Free fall problems are pretty simple, math and physics-wise, as long as you are certain of what values you have and what values you are looking for. You can solve any free fall problem through the use of one (or more) of four kinematic equations. *Kinematics* is the branch of physics that concerns itself with the motion of objects without worrying about the forces that got the motion going in the first place.

*A note – I still use the more general and classic version of the kinematics formulas with v, d, t, and a. The newer form of these, called suvat uses different letters (s, u, v, a and t) for the same variables. This is just a personal choice based on my teaching high school physics to students who are a little shaky on math and the idea of the formulas. I opt for the “one less thing to potentially confuse” rule. Don’t worry – by year 2, we’re moving into suvat. *

The four kinematic formulas that are used to describe motion and free fall are:

**x = v _{i}t + ½at^{2}**

**v _{f}^{2} = v_{i}^{2} + 2ax**

**v _{f} = v_{i} + at**

**x = (v _{i} + v_{f}/2)t**

For all the above:

x = displacement. The difference between the objects starting point and ending point.

v_{i} = initial velocity (the “from” velocity in a “from x m/s to y m/s” statement)

v_{f} = final velocity (the “to” velocity in the above)

t = time

a = acceleration – in free fall problems, that’s -9.8 m/s^{2}

To solve your problem, look for the formula that solves for what you’re looking for and has the variables that were given to you in the problem. It’s just that simple.

One very important note – find mass in the formulas above. I’ll wait.

Not there, is it?

The mass of the object does not affect the motion (speed, acceleration, velocity, travel time) of an object in motion or in free fall. Heavy objects *do not* fall faster than lighter ones. Everything falls at the same rate, which is to say, acceleration: -9.8 m/s^{2}. In free fall problems, acceleration is always -9.8 m/s^{2}. Don’t freak out if you have a problem and you’re not given an explicit value for acceleration. As long as you’re object is falling on earth, you’ve got it. If you’re not falling on earth, look it up.

I know what you’re thinking: what about a feather and a bowling ball? Right, right – the feather falls slower than the bowling ball of course. But why? There’s air between the objects and the ground, and both must push through that air – they both encounter air resistance, a “push back” from the air. That’s not anything that says they’re falling with different rates – that only means that the bowling ball is pushing through the air faster than the feather. As a result, it takes longer for the feather to reach the ground. If only you could take the air out of the picture and drop a feather and a bowling ball…

Or try it for yourself. Drop a pen and a flat sheet of paper from the same height at the same time. The paper takes longer to hit the ground, right? Air resistance – more for the flat sheet than the pen. Now, wad the paper up into a ball. The ball of paper and the sheet have the same mass, so that’s remained constant. Now drop the pen and the ball of paper. See?

In our example of Aquaman standing on a parademon and both of them in free fall? Barring being knocked to the side by high winds, Aquaman will continue to “stand” on the parademon all the way down. They’re both falling together at the same rate.

For what we’re going to be talking about here, we’re just going to ignore air resistance. It’s a cheat in a way but unless you’re talking about something with a large surface area and a lot of time spent falling (or just trying to get a larger point across), it’s a safe cheat. Your numbers will end up in the ballpark.

Oh, and there are some hard and fast rules:

- If the thing that’s falling is being dropped from its height, its initial (downward) velocity is zero. If something is thrown downward, or has been falling downward prior to us caring about it, then its initial velocity will not be zero.
- If something was thrown or projected upward along a perfectly vertical path, and it’s falling back to the earth, it will lose velocity as it goes up and at the top if its trajectory, its velocity will be 0 m/s. That’s the itial velocity for its downward trip. Side note – if you’re working the math for the trip up, the objects final velocity would be 0 m/s.
- Also, if the object was projected upward along a perfectly vertical path, it’s final velocity when it’s coming back to the ground will be equal (but opposite in direction) to the velocity with which it was shot upwards. 45 m/s going up? When it comes back down, it’s going to be moving at 45 m/s.

So let’s get back to Aquaman – he’s still falling with the parademon. As the formulas above tell us, the mass of Aquaman and the mass of the parademon don’t matter. The only things that matter are the velocities, the time and the displacement. As long as we know two, we can solve for the third.

For me and my students, this is the start of physics being magic.

With just a couple of bits of information, these formulas can tell you something that seemed to be unknowable.

Let’s take a look…

First off, to do any of these problems, we’re going to have to make some assumptions here and there – which is a good skill to nurture – and I’ll explain them as we go.

**Example 1: Aquaman is falling from 500 Feet. How fast is he moving when he hits the ground? **

I made that 500 feet up. Just giving a value there.

Step 1: Change those feet to meters. Physics, like all sciences uses the metric system. A quick conversion for 500 feet gives us 152.4 meters.

Step 2: What do we have, what do we need and what can we assume? We don’t have that much – a displacement (which is a distance and a direction, down) and acceleration, -9.8 m/s^{2}. That’s it.

What do we want to know? The speed at which the Aquaman/parademon combo is moving when it hits the ground, in other words, our final velocity, v_{f}.

We’re going to assume that he began his fall from a spot that had no vertical motion – maybe the pack of a plane, a high cliff…whatever. We’re just looking at the downward motion here. That means that the initial velocity was 0 m/s. As I said earlier, the initial of *anything* being dropped, rolling off a cliff, etc. is always going to be zero.

Again, we’re also going to assume no air resistance. It’s a cheat, but we’ll get by.

Step 3: What tool do we need? And by tool – we’re talking about our kinematic formulas. We’ve got acceleration, displacement, and an initial velocity of zero. Looking at the formulas, there’s one that will help us right out:

**v _{f}^{2} = v_{i}^{2} + 2ax**

Now it’s just plug and chug – put the values in and run the math.

v_{f}^{2} = 0 + 2(-9.8 m/s^{2})(-152.4 m) (the (-) in front of the displacement indicates it’s downward)

v_{f}^{2} = 2987 m^{2}/s^{2 }(this is the square of the final velocity – find the square root)

v_{f }= 54.7 m/s

But – what if the Aquaman/parademon combo has been falling for a little while, and we’re just catching up with them as the camera zooms in? Then we have an initial velocity, and the final velocity is still easy to calculate. Let’s say that they were moving at 25 m/s when we joined them at 152.4 m. How fast will they be moving when they hit the ground?

v_{f}^{2} = v_{i}^{2} + 2ax

v_{f}^{2} = 625m^{2}/s^{2} + 2987 m^{2}/s^{2 }(625 is 25^{2})

v_{f}^{2} = 3612 m^{2}/s^{2}

v_{f }= 60 m/s

Okay – a quick note on something reaching a speed of 60 m/s at a distance of 152.4 m – it may not happen. As I mentioned earlier, whenever something falls, air pushes back against it – think of sticking your hand out the window of a moving car. That air pushing against you slows you down. Eventually, if you’re falling through the air, you’ll reach a point where the air is pushing up with the same force with which you are pushing down (which is equal to your weight in Newtons).

At that point, you’ve reached terminal velocity. No, you don’t die, despite the ominous name. You just stop accelerating. Your downward velocity becomes constant, and you don’t speed up or slow down. A good ballpark value for a human(oid) in a prone position (like our parademon) is around 120 – 150 mph or somewhere between 54 and 67 m/s.

Once an object reaches terminal velocity, that’s it. How fast will it be moving when it hits the ground? Its terminal velocity. How fast will it be falling a minute and a half after reaching terminal velocity? It’s terminal velocity. Unless an object has some means of propulsion, it can’t fall faster than its terminal velocity, which depends upon the surface area of the object, the density of the air its falling through, and the object’s drag coefficient.

And remember, the speed at which something falls does not depend upon mass, so the formula above can be used for any object (there will be differences in the real world due to air resistance). By using the formula above, you can figure out the (rough) final speed of any falling object just by plugging a different value in for x.

**Example 2: Aquaman is falling from 500 Feet. How long until he hits the ground?**

Step 1: 500 feet = 152.4 meters.

Step 2: What do we have, what do we need and what can we assume? We’ve got acceleration (-9.8 m/s^{2}), we’ve got the displacement of 152.4 m.

We’re trying to find the time it takes Aquaman/parademon to hit the ground, and just to make things easy (we can adjust it later) we’re going to assume the initial velocity of the fall was zero.

Step 3: What tool do we need? With what we’ve got and our assumption, we’re need to find time. There’s no formula up there that solves directly for time, but there is one that has time as a variable as well as everything else we’ve got:

**x = v _{i}t + ½at^{2}**

And it’s just plug and chug time again:

-152.4 m = 0(t) + ½(-9.8 m/s^{2})(t^{2})

-152.4 m = 0 – 4.9 m/s^{2}(t^{2})

31.1 s^{2} = t^{2} (square root that thing)

t = 5.6 seconds.

And again, if we assume that Aquaman and the parademon had been falling for a while and had an initial speed of 25 m/s, that’s just going to decrease our time when we put the value into our original formula.

Still – all of this is independent of mass. Want to know how long it takes (roughly – again, we’re not worrying about air resistance here) to fall 200 meters? Put that in for x in the original formula. Remember to watch your signs – they’ll mess you up if you forget that down is negative.

Okay, okay – for both examples, I’m just talking about Aquaman and the parademon falling and hitting the ground. In the trailer, they land on the building and crash through all of it. What about that?

Honestly? I’d rather not.

Don’t get me wrong – there is some physics that can be done here. If we can correctly estimate the height at which the two fall we can subtract the height of the building from that, and then calculate the speed at which they’d be traveling when they hit.

If we know the speed, we can calculate the kinetic energy they’d have when they hit with the kinetic energy formula:

E_{k} = 1/2mv^{2}

So there’s the start of our problems – m – we’d need to assume a mass value for the two of them together.

And then there’s the crash through a five-story building (slow down the trailer and you can see the bodies in each floor) without slowing down. Moving through each floor eats up energy – it takes energy to break through the materials. You’d have to make assumptions about the tensile strength of the building materials. And then Aquaman pops out at the end and ruffles his hair?

Dude, that’s all just cool.

And would send us down a rabbit hole of complications and assumptions. Let’s just worry about freefalling objects.

If you want to, okay – think of it as homework – come up with reasonable estimates (that you can justify) for mass of Aquaman and the parademon, calculate the kinetic energy of them falling at a velocity of your choosing (calculated from a height of your choosing), and then compare that kinetic energy to energies of bombs. You should be able to find something comparable.

Let’s look at one last use of the kinematic formulas for free fall.

**Example 2a: How far would Aquaman fall in x seconds? **

Step 1: Nothing to convert.

Step 2: What are we given? Nothing, really. We know that acceleration will be -9.8 m/s^{2}, and we’re good to go.

What do we want to calculate? The time of a fall from any height we choose.

What do we assume? The height, which is the displacement that we’ll use in our formulas.

Step 3: Same tool, since this is just a modification of what we did in Example 2:

x = v_{i}t + ½at^{2}

Put any value in for x and solve for t. If you assume that Aquaman started his fall with no vertical motion, then you’re good to go. Or add in an initial velocity. Up to you.

Plug in any time you want, and plug and do the plugging and chugging. For example, how far would Aquaman/parademon (or, remember – anything) fall in six seconds?

x = v_{i}t + ½at^{2}

x = 0(t) + ½(-9.8 m/s^{2})(6 s)^{2}

x = -4.9 m/s^{2}(36 s^{2})

x = – 176.4 m (remember, the negative sign means down)

If you’re going to play with how far falling things fall in times of your choosing, just remember, that using this formula doesn’t give you a final velocity. That v_{f} value is important, because you don’t want to calculate that something falls farther than it would faster than it would because you forgot to consider terminal velocity.

Other applications of free fall calculations in movie and comic book style applications would be quite possibly two step problems where the terminal velocity is considered as v_{f1} and the terminal velocity is used for the second part of the fall.

So there we go – a falling Aquaman and pardemon combo, and a little physics. Now go and figure out final velocities, distances and times of falling things.