Determining altitude given velocities and mass

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Homework Help Overview

The problem involves a pole vaulter's motion, specifically determining her altitude as she crosses the bar while considering her initial and final velocities. The subject area relates to energy conservation principles in physics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of energy conservation equations, questioning the components of velocity and how they relate to height. There are attempts to clarify the vertical and horizontal components of velocity and their impact on calculations.

Discussion Status

Multiple participants have provided calculations and results, with some expressing uncertainty about their outcomes. There is a recognition of differing results, and participants are encouraged to check their math while referring back to the original equations.

Contextual Notes

Participants are working under the assumption that air resistance and energy absorption by the pole are negligible, which may influence their calculations. There is also a focus on ensuring that the correct components of velocity are used in the energy conservation equations.

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Homework Statement


A 52 kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.3 m/s. Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.


Homework Equations


1/2mv2 = mgy + 1/2mv2


The Attempt at a Solution


1/2 (52kg)(10 m/s)2 = 52kg (9.8 m/s2) y + 1/2 (52kg)(1.3 m/s2)
y=51.8m
 
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1.3 m/s is the horizontal component of her velocity. Initial velocity is given. Find the vertical component of the velocity and then find the height.
 
1.3^2 + b^2 = 10^2
b=9.91

the equation i used was:
1/2 (52kg)(10 m/s)2 = 52kg (9.8 m/s2) y + 1/2 (52kg)(9.91 m/s2)
y=0.091m
 
Your initial equation is correct and it will give you 4.935m as the height for which she jumps. When looking into conservation of energy you do not deal with vectors only magnitudes.
 
how did you get 4.935m?
 
I just solved the equation you had in the beginning. I am pretty sure the method is correct, check your math.
1/2 (52kg)(10 m/s)2 = 52kg (9.8 m/s2) y + 1/2 (52kg)(1.3 m/s)^2, solve for y.
 
hmm, i must have put something in wrong. 4.935 was correct. Thanks
 
hmm I can't get 4.95..

i'm getting 5.01..

but your initial eq'n is right.
 
josh, i kept getting 5.01 as well
 

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