Motorbike Bridge Crash [LONGER VERSION]

Lots of questions about the crash, folks asking for more details etc. A number of folks asking for a longer version. Here it is.
Video Rating: 4 / 5

Ghost rider. I wouldn’t try this on the streets. Awesome bike though. Turbo hayabusa

Question by Nick McKay: Motion in two Dimensions, the speed required to jump a motorbike?
A stint driver on a motorbike wishes to shoot of a 35degree incline and land on a platform 15m in the x direction and 2m high (above the top of the incline, its y displacement is 2m up). How fast must the motorcycle be moving if the stunt is to suceed?

I thought, ‘if it only needs to suceed, than its lands on the platform in the y direciton with a velocity of 0, but it didn’t work out. Thank you very much.

Best answer:

Answer by zee_prime
Right. The vertical component of his speed is v sin 35 and the horizontal component is v cos 35. He will be airborne for t seconds. It will take him 15/(v cos 35) seconds to cross to the platform. At that time he must be at least 2 m high. His height is given by the formula h=vt +4.9(t^2) where g = 9.8. So, substituting for cos 35, t=18.31/v. Plugging this into the second formula, h=18.31+4.9 (335.3/v^2) so 2=18.31 +1643/v^2 so 16.31+1643/v^2=0 so v^2= 1643/16.31=101 so v must be at least 10 metres/sec if he wants to stay on this side of the river Styx. If I was him, I’d allow for air resistance etc and go a bit faster.

Know better? Leave your own answer in the comments!

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