My Math Forum why we use this formula

 Physics Physics Forum

 December 23rd, 2010, 01:11 PM #1 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 why we use this formula A 50-kg 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. $E_i=E_f$ $K_i+U_i=K_f+U_f$ $\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$ $h_i=0$ and $h_f=h$ thus: $\frac{1}{2}mv_i^2=\frac{1}{2}mv_f^2+mgh$
 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.

 Tags formula

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# A 54.6Â kgÂ pole vaulter running at 11.9Â m/sÂ vaults over the bar. Her speed when she is over the bar is 1.38Â m/s. Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.Â

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post NuNu_dagobah Computer Science 4 January 8th, 2013 07:31 AM nsrmsm Number Theory 7 October 15th, 2012 05:36 AM djx18 Algebra 1 September 20th, 2009 05:25 PM agro Probability and Statistics 3 August 27th, 2009 06:17 AM NuNu_dagobah Abstract Algebra 0 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top