More on Craig’s Recent “Take a Cool Guess” Post
Reader Drew Douthit adds this clarification to my post on the drawing here: “Take a Cool Guess—The Fun Quiz on Renewable Energy and Environmental Sustainability. Today’s Topic: Basic Physics”
It all depends on vehicle speed….
okay! For 165lbs diver it’s over 67mph!
Note that the cartoon is intentionally misleading for the purposes of humour. It’s drawn as if the diver had no forward velocity, so as soon as he jumps the vehicle will shoot away from underneath him. However, in the real world, at the moment of taking off the diver would have a forward velocity equal to the velocity of the vehicle. So in real world situations the diver could land in the pool.
See [this amazing video](https://www.reddit.com/…/can_someone_explain_the…/)
of a motorcycle passenger going airborne at speed and landing in her seat.
But the situation in the cartoon could happen, if the vehicle was going fast enough and/or there was a strong enough headwind. In this scenario, although the diver starts his dive with forward velocity, the air/wind resistance will reduce that forward velocity and he could miss the pool as the vehicle continues to move with constant velocity.
Let’s try to do the maths. We’ll need to make some estimates and assumptions.
In the picture I’d estimate that **the diving board is about 1 metre high, and about 2 metres long** from the take-off point to the back of the vehicle.
I’ve watched some videos of Olympic platform divers and they don’t tend to gain much height as they take off. Yeah, the guy in the cartoon is on a springboard, but it’s not a very long springboard and statistically he’s unlikely to be an Olympic class diver. Let’s assume he gets about **1 metre above the board** at the peak of his dive.
We can easily use the Newtonian equations of motion (**s=½at^(2)** etc) to calculate how long he’s in the air. With a = -9.8 m/s^2 we find that jumping 1 m takes 0.45 s and then falling 2 m takes 0.64 s, for a **total air time of 1.09 s**.
When we’re looking at the horizontal motion, we can use the vehicle as an inertial reference frame. So we have the diver starting off stationary, and then when he jumps he’s subject to a horizontal force due to the air. Note that we won’t be able to distinguish between vehicle speed and wind speed: jumping off a stationary vehicle into a 50 km/h wind is the same as jumping off a vehicle moving at 50 km/h through still air. Anyway the question is what wind speed does the diver have to experience, for it to move him 2 m horizontally, from a standing start, in 1.09 s. If we assume that he’s subjected to a constant acceleration, Newton’s equations tell us that the **acceleration must be 3.36 m/s^(2)**.
Newton tells us that **F = ma**, and the force of air resistance is **F = ½CpAv^(2)**, where C is the drag coefficient, A is the cross sectional area, and p is the density of the air. So we can set these two expressions equal to one another and then rearrange to get **v = √(2ma/CpA)**.
We calculated a = 3.36 m/s^(2), we’ll assume the **mass of the diver is m = 75 kg**, and we know that the density of air (at sea level and 15°C, the International Standard Atmosphere) is p = 1.225 kg/m^(3). Though having said that, if the air temperature were 15°C, I don’t think we’d be going swimming and diving. Let’s assume the air temperature is 25°C, in which case the **air density is p = 1.168 kg/m^(3)**. So we plug all these values into v = √(2ma/CpA) and we have **v = √(431.87/CA) m/s**.
Now we need values for the drag coefficient C and the diver’s cross sectional area A. Fortunately I found [this source](https://phys.libretexts.org/…/6.07%3A_Drag_Force_and…) which states that a human skydiver has a cross sectional area of about 0.18 m^2 and a drag coefficient of about 0.7 in the head down position, and a cross sectional area of about 0.70 m^2 and a drag coefficient of about 1.0 in the spread eagle position.
Now this is a bit awkward because obviously a diver changes his orientation, and hence his cross sectional area and drag coefficient, during the course of his dive. But, looking at the cartoon, in the second panel the diver hasn’t really had enough time to pivot into a head down position. He looks like he’s been pretty close to the spread eagle position throughout. So I’m going to take **A = 0.70 m^2 and C = 1.0**. That also means I don’t need to worry too much about that assumption of constant acceleration.
And here’s our answer! We have v = √(431.87/CA) = 24.8 m/s which is **89 km/h or 56 mph**. As noted previously, that’s the speed of the air into which the diver is jumping. It could be a vehicle moving at 89 km/h in still air, or a stationary vehicle in an 89 km/h headwind, or a vehicle moving at 120 km/h with a 31 km/h tailwind, or any other combination that adds up to 89 km/h.
**EDIT:** If you’d like to calculate your own answer, you’ll need values for the following parameters:
* m = mass of diver
* l = length of springboard
* g = gravitational acceleration
* C = diver’s drag coefficient
* p = density of air
* A = diver’s cross sectional area
* j = height of diver’s jump above board
* h = height of springboard above vehicle
and then plug them into the following equation:
**v = √(2mlg/CpA) / ( √j + √(j+h) )**
Good luck trying that with anything other than SI units!
**EDIT 2:** It just occurred to me that I implicitly assumed the diver would jump slightly sideways, so that he wouldn’t land on the springboard as he came down. That’s perhaps not realistic. But if he jumped straight up, the situation is that he has to travel the 2 metres back towards the end of the vehicle before he comes back down to the level of the springboard. This is easily achieved by setting **h=0** in all the calculations above. We get v = 30.0 m/s which is **108 km/h or 67 mph**.
