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$h=1.4m$

Giancoli 6th Edition, Chapter 6, Problem 33

(2:12)

Chapter 6, Problem 33 is solved.

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Transcript for this Giancoli solution

The size of Jane's change in kinetic energy is going to be equal to the amount of change in potential energy. I would like to use this absolute value signs just so that we don’t have to worry about negative signs, it's very annoying. So we have, the change in kinetic energy is going to be all the kinetic energy that Jane started with because she's going to end up stopping. And then the change in potential is 'mgh' and we'll solve for 'h'. So divide by 'mg' , the 'm's' will end up canceling which is nice and so we have 'h' equals 'v' initial squared divided by '2g' which is 5.3 meters per second we'll square that. Divided by 2 times 9.8 meters per second squared. And we get ANSWER meters is how far Jane goes up on the vine.

In the fifth edition she's actually running at 5.6 meters per second so we'll square that and in the fifth edition the answer will instead be ANSWER meters. That’s for part A.

And in B it asks will this height depend on the length of the vine? And the answer is, no it wont. So go up this far regardless of the length of the vine. The only thing that would change is the angle that would be made with respect to vertical. So on a short vine she would end up having a large angle. You know this is ANSWER meters going up whereas on a long vine you can imagine going up the same height of ANSWER meters but in this case theta. Theta long against theta short, theta long will be a lot less than theta short.

Comments

Cheers, thought it was gonna be harder than this, once again, making it 10x easier. I look forward to sharing a brewski with you one day. You're the best.

Hi Oliver,

I'm really glad the videos help! Australia's a long way from Canada, but I like your thinking. :)