I made a gamble and my gamble failed. It really wasn't my fault. In the preview, they showed this huge barrel thing dropping on a see saw. It looked something like this: (I could search for 30 minutes for a picture of this online, or just draw it)
At first glance, this looked JUST like that video of the pile driver shooting the skydiver up in the air. You can see how I would make that mistake. To make up for my mistake, I will give a very simple analysis of the see saw myth. The basic idea is that something comes down, hits the see saw and sends the other thing flying up. If I assume there are not energy losses and take the two objects plus the see saw as the system and Earth, then there is no work done so that:
Where E1 and E2 are two energies at any time. If I take E1 to be right before the sky diver hits the see saw and E2 to be when the child on the see saw gets as high as possible then I can write the energy equation as: (note, I am denoting the mass and speed of the skydiver with a subscript 's' and the child with a subscript 'c')
Here I am letting the ground level be the zero gravitational potential energy (though it doesn't really matter where you put it). So, right before the collision, the skydiver just has kinetic energy and the child has nothing. After the child gets thrown up into the air and at the highest point, all the energy is just in the gravitational potential energy of the child-Earth (child is not moving at highest point). I could solve this for the height, and I get:
Let me go ahead and put in the values they used in the show. I am pretty sure they had the skydiver at 170 lbs going 122 mph (54.5 m/s) and the child was 50 lbs. I don't need to convert the mass of the two objects since they will cancel. This gives a height of 515 meters or 1689 feet.
Remember, that assumes ALL of the energy of the skydiver went to the child.
- They claimed a skydiver would need to fall 600 feet to reach terminal velocity, is this true?
- If you wanted to drop something from a lower height, what would its mass have to be to be equivalent to a 170 lb skydiver going 122 mph?
- Suppose you want to launch a skydiver straight down with a spring (bungee cord). If you shooting down from 100 feet, and you stretch the cord 50 feet, what would the spring constant need to be to get the skydiver to 122 mph?
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Those are really interesting questions. I was just happy to watch something that had some intelligence on tv.
Ideally (in the modeling sense) the skydiver never reaches terminal velocity, he or she approaches it asymptotically. For another Mythbuster episode, I made a spreadsheet to model the falling of a skydiver and searched for consensus numbers for skydiver cross sectional area and drag coefficient. For the first pass, I assumed constant air density, then I did another using increasing air density as the diver falls. I posted the results on the Mythbuster fan site.
In any case, the skydiver requires on the order of 900 feet to reach 95% of a terminal velocity on the order of 90 miles/hour. This assumes the diver assumes the spreadeagled position taught in skydiving school.
I'm not sure why they would call this one definitively busted. Sure, the girl went too high and it killed her, but they built a near-ideal see saw that imparted nearly all of the force to the girl. On their first attempt, the girl did not go high enough because the see saw bent and thus did not send all of the energy to the girl. So, isn't it reasonable to assume that if you had a slightly weaker see saw (than the ideal one they made) that you could get the girl the correct height? Also, maybe that amount of force wouldn't kill her. It might, but they didn't really test that. I think this one needs to get revisited.