December 23rd, 2010, 01:11 PM  #1 
Senior Member Joined: Oct 2009 Posts: 895 Thanks: 1  why we use this formula
A 50kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.0 m/s...? Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar. In my book the soving start by (1/2)mvi² = (1/2)mvf² + mgh please help me .. 
December 23rd, 2010, 01:24 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: why we use this formula
Your book is equating total initial energy to total final energy. and thus: 
December 23rd, 2010, 08:42 PM  #3 
Senior Member Joined: Oct 2009 Posts: 895 Thanks: 1  Re: why we use this formula
But we use here (total initial energy to total final energy) tell my the concept about this Q

December 23rd, 2010, 08:54 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: why we use this formula
Well, as I said before, we are setting the total initial energy equal to the total final energy. Just as the vaulter leaves the ground, all of her energy is kinetic, or relating to motion, then during the vaulting process, some of her kinetic energy is converted into gravitational potential energy. While she now has two forms of energy, the sum total remains constant. As the kinetic energy is converted into potential energy, she slows down. In the problem, her negative change in kinetic energy is countered by a positive increase in potential energy. It is through the fact that her total energy remains constant that we can find her altitude, knowing her initial and final velocities. This is the same fact that we use in a variety of problems involving changes in kinetic and potential energies. 

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