How to calculate entry speed in high diving

In high diving, divers jump from platforms of varying heights, and the higher you go, the faster you will be going when you hit the water. I've seen many people cite different specific speeds for different heights, and at least a couple of mistakes, so here is an explanation of how this speed can be calculated.

The formula

If you're just here for the answer to the question in the title, the formula is this:

Where v_entry is the velocity of a diver when they hit the water in m/s, g is gravitational acceleration, g = 9.80665 m/s² (this value varies slightly across the globe, more on that later), and h_total is the highest point a diver reaches during a dive (for example, if the platform is 12 meters above water, and the diver jumps 30 cm up, h_total is equal to 12.3 meters). If you want the result in km/h, multiply the result by 3.6. For miles per hour, multiply by 2.24 instead.

As an example, if a diver jumps from a platform  that is 27 meters high, and jumps 30 cm up, the result is:

This means that at the point when the diver touches the surface of the water, he will be travelling at 23.14 meters per second. Multiplying that number by 3.6 gives us 83.30 km/h, and similarly, multiplying by 2.24 gives 51.83 miles per hour. This equation is not considering air resistance, but is precise enough to work with. If you are not satisfied with neglecting air resistance, read on.

The math and physics behind

To understand where this formula comes from, we must first understand a bit about the kinematics equations, which describe the acceleration, velocity (or speed), and displacement of an object over time. Specifically, we will take a look at the equations describing an object in freefall.

Imagine yourself on the edge of your favorite 10 meter platform, holding in your hand some object, say a tennis ball, over the water, ready to drop. The exact moment you let go of it, it will have no velocity, and then from there it will start to gain speed, as time passes. The fact that the speed of the object changes over time means it is accelerating, that is what acceleration is. This is the reason an acceleration such as gravitational acceleration has the units [m/s²], which can be read as meters per second per second. It means that the speed (measured in meters per second) changes by some amount every second, thus the twofold 'per second'.

For the ball we just dropped, it will start accelerating downwards, towards the Earth's center of mass. As it gains speed, it will also experience a force accelerating it in the reverse direction (or decelerating), caused by air resistance. In the above equation this is neglected as mentioned, and I will talk about the arguments for this a bit later. So for now, let's assume no air resistance. In this case, the only force left is gravity.

To be precise, the gravitational acceleration, g, varies depending on how far you are from the Earth's center of mass. This means that the acceleration is lower on mountains, and near the equator (due to the Earth bulging at the middle). On the surface of Earth, the lowest measured value is g = 9.7639 m/s², on the surface of the Arctic ocean, and the largest is g = 9.8337 m/s², at the peak of the Nevado Huscaran mountain in Peru. These are absolute extremes, and still the value varies less than a percent. Therefore, we simply use a standard value for g, which is the one mentioned above, as it will not impact the calculations significantly. This also means that the acceleration of the ball in the equations is constant, which means that it does not change with time. This makes the equations much simpler. Note also that the gravitational acceleration is independent of mass, which means that heavier (or denser) objects accelerate at the same rate as lighter (or less dense) objects. This is only strictly true when there is no air resistance, as famously demonstrated by astronaut David Scott on the surface of the moon, which has no atmosphere, but as mentioned, we are assuming air resistance to be negligible.

Another thing to note is that we only considering a single spatial dimension in these cases, the up-down direction. This means that we are not considering any side-to-side, or forwards and backwards motion, and when we will later talk about jump height, we are also only considering the vertical jump height. 

As mentioned, the things we need to know about are acceleration, velocity, and displacement. Since we consider gravity to be the only acceleration on the ball, the equation for acceleration is simply:

Notice that there is no t in the right hand side of the equation, which signifies that the value doesn't change with time.

As mentioned, an acceleration determines how much an object changes its velocity over time. For example, if an object is accelerating at a rate of 1 m/s², and starts at rest, it will be traveling at 1 m/s after one second, 2 m/s after 2 seconds, and so forth. The general formula for this is:

Where a is acceleration, t is time, and v_0 is starting velocity, if any. In the case of our ball (and later a diver), the acceleration is simply g, and we start with no velocity (holding the ball still in our hand), which means v_0 = 0, so the specific equation for our case is simply:

The last equation is for displacement. Displacement means simply how much an object has moved. If an object, say a meteor, is travelling through deep space at a constant speed of 10 km/s, after one second it will have travelled (its displacement will have increased by) 10 km, after two seconds, 20 km, and so forth. In the case of our ball however, its speed is not constant, it is changing over time. Specifically, it is changing at a constant rate. In this specific case of constant change of velocity, the displacement equation is:

Where a is the acceleration of the object, t is time, v_0 is the starting velocity of the object, and s_0 is the initial displacement of the object.

For those familiar with calculus, they may have noticed that this is simply the integral of the velocity equation above, which in turn is the integral of the acceleration equation. If you are not familiar with calculus, this equation probably doesn't make intuitive sense. It can be thought of as the sum of velocity over time, which becomes displacement. 

In any case, in our case with the ball being dropped, we have established that v_0 is 0, and a = g, so they can simply be plugged into the equation. For s_0, lets also set this to 0. Then we will have an equation that describes how much the ball has moved over time. This gives us our equation for displacement in the case of the ball:

Now that we have equations for acceleration, velocity, and displacement over time, we can plot them, to see what happens to the ball as it falls.

We know that at some time, let's call it t_entry, the ball will hit the water, because it will have travelled 11.5 meters downward. If we plug the value -11.5 meters into the displacement equation (negative because thats the downward direction), we suddenly have an equation with only one unknown, which can be solved for t_entry. The equation is:

Note that i have also denoted g as negative here, as explained above, though it is the same if both g and 11.5 were positive (try multiplying both sides by -1).

From this, we can isolate t_entry like such:

Plugging in g = 9.80665 m/s², we get t_entry = 1.5315 s, which means that the ball hits the water 1.5315 seconds after it has been dropped from a height of 11.5 meters.

We can now take this value of t_entry, and plug it into the equation for velocity, to get how fast the ball is travelling at that time:

This gives us the answer, the ball is travelling at 15.018 m/s, the instant it touches the water (54.07 km/h or 34.64 mph).

So to reiterate, we plugged a certain height into the displacement equation, and solved for t_entry, to get the time it takes to fall that height, and then we plugged that t_entry into the velocity equation to get the velocity at impact with the water. To generalise, we can find t_entry for any height h_total, using this formula:

And multiplying this by g (remember v(t) = g*t), we get the final formula for calculating entry velocity:

Using a bit of algebra, we can simplify this term a bit, arriving at the formula presented at the beginning:

So there it is, simply input the platform height + jump height, and use your favorite value of g (just use 9.80665 m/s²), and you have the velocity at entry. Plotting v_entry, against h_total shows the entry speed at a range of heights (multiplied by 3.6 to get result in km/h):

The graph shows that as the platforms get higher and higher, each meter adds less and less speed to the entry. This is because of the square root in the formula, and it is called "sublinear growth" (as opposed to linear or exponential growth for example).

As a small footnote, these velocities are not strictly speaking the maximum velocities during the dive, only the velocity as the feet of the diver touches the water (assuming a standing dive; for handstand dives this statement is slightly incorrect). When a diver hits the water, they will very quickly start decellerating, but the increase in decelleration (the jerk) isn't instant, which means there must be some very small amount of time where the decelleration is less than g, which means the diver is still accelerating downwards, increasing velocity. I can not say whether this is milliseconds or nanoseconds, but it must be there.

Jump height

So far we have just been using 30 cm as a standard value for jump height. This value is from this article, which studies the hydrodynamics of a 10 meter straight jump performed by Dave Colturi. The entry velocity of a diver does not change significantly with jump height at higher platforms. For example, the difference in entry velocity of jump heights 10 cm (very low) and 100 cm (very high) from 27 meters is only 1.37 km/h (82.99 km/h vs. 84.36 km/h).

There are however some reasons to elaborate further on jump height; first of all, if you want to know really precisely what velocity you hit the water with, you will be interested in your precise jump height. Secondly, the amount of time you are in the air (air time), will differ based on your jump height. Specifically, if you want to know how much relative extra air time you have going from 20 to 27 meters for example, you have to know your total airtime for both, not only the increase (which can be found using simply the formula for t_entry above).

I will write about how to find your jump height, and how it affects your airtime in the next blog post.

Air resistance

So far we have been neglecting air resistance, or 'drag'. Drag is a force that slows down an object that's traveling through a fluid (or vice versa). In our case we have an object (a diver) traveling through a fluid (air), with increasing velocity. In the process of falling, the diver will come to occupy space that was previously occupied by air. The diver needs to push this air away, which requires some amount of energy, which will ultimately reduce the velocity of the diver.

The problem with all this is that it relates to fluid dynamics which is very annoying to deal with (for me anyways), and not nearly as simple as the above equations of motion. Let's scratch the surface.

In the case of drag, fluid dynamics describes two versions of drag, linear, and quadratic, which simply means one increases linearly with velocity, and the other quadratically (with velocity squared). In many cases, one force dominates, and to examine that, fluid dynamics uses a quantity called Reynolds number: 

Where Re is the Reynolds number, ρ is the density of the fluid [kg/m^3], u is the flow speed [m/s], L is the characteristic length, and μ is the dynamic viscosity of the fluid [kg/(m*s)].

Note that the Reynolds number changes with flow speed, which in this case is the velocity of the diver, which means it is not constant for our situation.

The linear drag is dominant in situations when Reynolds number is much smaller than 1, and if we plug in some values, we get the change of Reynolds number over the course of a dive form 27 meters:

Here I have used 1.225 kg/m^3 for the density of air, 3.178*10^-5 for the dynamic viscosity of air, and 20 centimeters for the characteristic length. The characteristic length is a complicated subject, so I have chosen a rather small value to be on the safe side.

Since the Reynolds number is far above 1 for almost the entire duration of the dive, we will be neglecting the linear drag force, and only focusing on the quadratic drag force.

The equation for quadratic drag is:

Where F_d is the drag force [N], ρ is the density of the fluid [kg/m^3], v is flow velocity [m/s] (convention is to use u, but i have used v for consistency), c_d is a drag coefficient [dimensionless], and A is the reference area [m^2].

Here it can be seen that the drag force is quadratically proportional to velocity, since the v is squared.

Before we get into it, lets quickly look at these variables. The fluid density ρ which we also used for the Reynolds number is quite simple, and is simply the mass of air, by its volume. The velocity v is called flow velocity, which might be confusing in the case of a dive, since the air is standing still and the diver is moving. This is however equivalent the the diver being stationary, and the fluid around him accelerating, so if we imagine it like this, the flow velocity term makes sense. The drag coefficient c_d is also a bit difficult, but for a human falling, it is around 1 belly down, and around 0.6-0.7 falling feet first. The reference area A can in our case be described as the area of the divers shadow, when light directly from above. Say you're in a tuck position, your shadow will approximate a circle. 

Now finally, we need to talk about what a force is; a force is mass times acceleration, as per Newtons second law:

Since what we are actually interested in is entry velocity, we want to know how much the drag forces changes the velocity of the diver, which is acceleration. According to the second law, acceleration is force divided by mass, so to get the acceleration due to the drag force, we have to divide it by the mass of the object, the diver:

Now if you remember back to the start of this discussion, you would remember that velocity was a function of acceleration (v = a*t), but now we have an acceleration that is a function of velocity. That means we are dealing with a differential equation, which makes me significantly sadder. Fortunately, much smarter people than me have solved the problem, and have derived equations for both velocity and displacement, taking quadratic drag into consideration. The formulas are:

For velocity, and

For displacement, where

The Wikipedia article for hyperbolic functions has the answers to some of your questions (not all).

With these, we can now plug in some values, and compare them to our initial calculations, to see how much air resistance matters, and whether or not it is reasonable to neglect it.

To start, let's take an extreme example with a lot of air resistance; a parachute dive. If Google is to be trusted, a parachute has a terminal velocity of around 28 km/h (7.78 m/s). Since the terminal velocity of an object is related to its k value by

we can find a value for k (in this case assuming the parachuteer + parachute weighs 80 kg), which is just around 13 kg/m (don't worry about the unit). Plugging this value into the velocity and displacement equations, with and without considering air resistance, we can see the difference:

The velocity graph shows that without considering air resistance, the velocity rises linearly as normal, but when we consider air resistance, the velocity flattens out, approaching the terminal velocity. On the displacement graph, we can also see that the air resistance consideration causes a much slower travel. It shows that while not considering air resistance, it takes 1.43 seconds to travel 10 meters, while it takes 1.83 seconds with air resistance. Plugging these two values into the velocity graph tells us that the velocity after travelling 10 meters is 14.0 m/s without air resistance, and 7.6 m/s with air resistance.

This shows us at least that the two equations are showing different results, and it also shows that air resistance can not be neglected when making predictions about parachute jumping, unless you are falling less than two or three meters.

Next, let's look at another example - a skydiver. According to this air resistance calculator, a skydiver has a k value of 0.5. This corresponds to a terminal velocity of 142.6 km/h (39.6 m/s), which is a bit slower than what Google says; according to the first result on Google, a skydiver falling stomach first has a terminal velocity of around 120 mph or 193 km/h, which corresponds to a k value of around 0.27. But for now, let's use the larger value of 0.5 (remember a larger k value means object is more affected by air resistance). Using this k (and still a mass of 80 kg) we get:

In this case, the terminal velocity of 39.6 m/s is still well out of range, but the air resistance of the skydiver is still slowing them down. Here, it takes 2.35 seconds and 2.41 seconds to travel 27 meters, without and with air resistance respectively, resulting in velocities of 23.0 m/s and 21.1 m/s (82.8 km/h and 76.3 km/h). 

So now, let's try to get as close as possible to the reality of a high diver. Let's try to calculate k for a diver, using the formula above. For that we need air density, ρ, a drag coefficient c_d, and a reference area A.

Air density can vary, but for our example, let's use the standard value for air densite at sea level, ρ_air = 1.225 kg/m^3. The drag coefficient is still a bit difficult, it depends on the geometry of an object, and also in which direction it is flowing (for example it is larger for a pipe falling sideways than the same pipe falling end first). As mentioned earlier, the drag coefficient for a human is around 1 falling feet first, and 0.6-0.7 falling belly first. Since most dives are rotating, let's use an average value of 0.825.

That leaves the reference area - as mentioned, when considering a falling object, it can be thought of as the objects shadow when lit directly from above (opposite the direction of gravity). If we were talking about something travelling in another way, like a car on a road, this analogy breaks down, but for now its usable. During a high dive, our shadow changes based on our position (tuck, pike, straight), as well as our rotation. In a tuck, our shadow is approximately a circle at all times, while in a straight position, it is somwhat a rectangle, shifting length as we rotate. Taking our arms out to the side, like some barani techniques, also increases the shadow area - it also increases turbulence because of the more uneven shape, but that is far out of our scope.

If we assume a tuck position to have a reference area (shadow) similar to a circle, we simply need a radius R to calculate it (A = πR²). Using my own measurements as an example, the circle has a radius of around 35 centimeters, resulting in an area of 0.385 m². For the straight positions rectangularly shadow, we can use a height and a width. If we use the distance from shoulder to shoulder times height of the diver, we get the largest value for the rectangle, which can serve as the extreme case. In my case this is 45 cm times 180 cm (don't @ me), which is 0.81 m². If we use the distance between shoulders and the depth of chest, we get the shadow of a diver travelling feet first in a straight position, which can serve as the minimum case. For me it is 45 cm times circa 35 centimers, equaling 0.16 m². The average of the two values for the straight position is 0.48 m², and the average of this and the value for the tuck is 0.43 m². Let's use this as a close-to-reality value.

Using these values, we get k = 0.219, so let's plug this into the formulas (still using 80 kg for mass):

Now we see almost no discernible difference in the displacement graph; the time it takes to travel 27 meters only differs by 0.027 seconds. The resulting velocities are 82.8 km/h, and 79.9 km/h, a difference of 3.6%. If we look at 20 meters, the difference in time is 0.019 seconds, and the velocities are 71.3 km/h, and 69.4 km/h, a difference of 2.7%.

While the differences in travel time are in my opinion extremely negligible, the difference in entry velocity is a bit larger, especially at 27 meters (and beyond). In my opinion the difference is not so large that the simple equation is useless, but I would use its results to very many decimal places.

So if you want to be precise in your velocity calculations, use the equations for velocity shown above. To use it, you will need to calculate a value for k, which you can do like I showed above, as well as a mass of the diver considered. To know the entry velocity from a specific height, you simply need to input the time t_entry into the velocity formula. As mentioned, travel time really isn't very different from the non-air-resistance equation, for low k-values, so you can use the simple formula for t_entry (square root of 2*height divided by g).

However, if you want to be really precise, you can use the value for t_entry with air resistance, which is:

And just to reiterate, this gives us this formula to calculate entry velocity with air resistance:

Conclusion

The entry velocity of a high dive changes with the height of the dive, this much was always certain, and now we know a bit more about the specifics. As with all physics, these calculations will remain approximations, so the only real question to consider is how much precision you need for your use case. Now, at least, you have the possibility to choose between different grades of accuracy for this specific problem.