Tarzan & Jane
Posted by jonathanshaffer1969 on 17th March 2010
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IMDB rating: 4.60 Plot: As Tarzan and Jane’s one-year wedding anniversary approaches, Jane searches the jungle for the perfect gift for Tarzan, even enlisting the help of Terk and Tantor. As they recall the many adventures they’ve shared so far, Jane realizes what an exciting year it’s been in the jungle from encounters with old friends, outsmarting panthers to surfing the lava down an erupting volcano. But all of that is nothing compared to the surprise that Tarzan has in store for Jane. |
Actors: Ellis Greg,Auberjonois Rene,Bennett Jeff,Cummings Jim,Denisof Alexis,O’Hurley John,Proctor Phil,Richardson Kevin Michael,Weiss Michael T.,Animation,Family,
Calculating height in physics?
Hello, does anyone know how to solve this physics problem? And can you explain it as well?
1. Jane, looking for Tarzan, is running at top speed (5.6 m/s) and grabs a vine hanging vertically from a tall tree in the jungle. How high can she swing upward?
This problem is one of conservation of energy. In the initial state all of the energy is in Jane’s motion, and in the final state it is all stored as gravitational potential.
Thus, it is simply a matter of setting the two forms of energy equal to each other:
1/2 m v^2 = m * g * h
In this case, as is common in conservation of energy the mass of the object (Jane, in this case) is irrelevant, since it will cancel out. Thus:
1/2 v^2 = g h
h = v^2/(2g)
With numbers, h = (5.6)^2/(2*9.81) = 1.598m
Hope this helps
Not always right (but usually) | Dec 17, 2009
this is a problem of conservation of energy. At the top of her ascent she will have zero kinetic energy, all Potential.
Following U+K=U+K
1/2mv^2=mgh
1/2v^2=gh
15.68=gh g=9.81 m/s^2
h=1.6 m
BTW this problem can probably also be done through kinematics
Steved | Dec 17, 2009
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