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8

Assuming terminal velocity of 200 km/hour, the scenario seems equivalent to stepping out of a car that's travelling at 200 km/hour. In that case it's not the fall (hitting the road) that kills you, it's the friction (i.e. sliding or tumbling along the road). There might be a minimum of friction initially (when you're falling parallel to a vertical wall) but ...


14

I have slid down a much smaller version of this at Burning Man. Paha'oha'o was a 30 foot tall volcano art piece which you climbed and then "sacrificed" yourself by dropping into a pit featuring a slide just like you mention. The drop features a 10 foot free-fall, just enough to take your breath away, after which the careful curve of the slide gently catches ...


7

Probably the closest to what you are asking about is the story of Ivan Chisov's survival (see Ivan Chisov); but there have been several other similar cases (see for example 10 Amazing Free Fall Survivors).


35

The answer is Yes and your thinking is correct. You try to differ between impact and sliding on a curve. In fact the impact is just a sudden large force, while a curved (e.i. circular) motion similarly applies a force, just much smaller but also over a longer period of time. The key in surviving any fall is to reduce the force on your body at "impact". A ...


26

Let's make life easy for ourselves by assuming that the slide is an arc of a circle: We also assume the slide is made out of something with a very low friction, so the skydiver maintains a constant speed $v$ all the way round. The reason that using an arc of a circle makes life easy is that the acceleration felt by the skydiver is simply: $$ a = ...


2

Yes. In fact it would be better to imagine that you skydive towards a "track" that you can strap a "chair" onto, and then the chair is stuck on the track. the acceleration to keep you in a circular orbit of radius $R$ is only $v^2 / R;$ with terminal velocity being about $v \approx 56 \text{ m/s}$ a $1~g$ acceleration will be accomplished by a radius of ...


0

a = 1.25m/s^2 Now , v = u + at = 1.25 x 8 = 10 m/s Hence s = 1/2 at^2 = (1.25)(64)/2 = 40 m Now, H= -40m ( due to downward direction.) and 40 = 10t - 5t^2 ( s= ut +1/2 at^2)(placing above obtained and given values) therefore , t^2 - 2t + 8 = 0 **t = 4**


0

It is because general theory of relativity. As it says, one person is in a lift suddenly someone cut the cable, then he will experience no gravity as both the lift and person are falling at same rate.



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