Free fall arrest

  • #26
Bandersnatch
Science Advisor
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Also, v0 is initial velocity, it should be zero. You put (54*3) as v0*t.
No, David has it right.
It shouldn't be 0. It is the velocity at the top of the contraption, i.e. when the person is falling at full speed and begins decelerating.
The way you wrote it, you will get the same numerical result, but with the opposite sign, and its meaning will be different:
Sign convention is + means up, - means down.
Since 0 is the ground level, ##r_0## with a negative sign means below the ground level. The meaning of the equation the way you decided to write it is:
The skydiver has 0 velocity at some distance ##r_0## below ground level and begins accelerating upwards until he reaches the surface, at which point his final velocity is ##V=at##, i.e. 54 m/s.
It's more of a catapult than an arrester this way. :wink:
 
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  • #27
No, David has it right.
It shouldn't be 0. It is the velocity at the top of the contraption, i.e. when the person is falling at full speed and begins decelerating.
The way you wrote it, you will get the same numerical result, but with the opposite sign, and its meaning will be different:
Oh, you're right. Duh moment.

Well in that case, v0*t should equal (.5*at2)*2 so the equation can be simplified to
r0=v0*t/2, or something like that.
 

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