Maximum height reached by a projectile

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SUMMARY

The discussion focuses on calculating the maximum height reached by a projectile, emphasizing the integration of velocity functions. The user initially considers equating kinetic energy (KE) and potential energy (PE) but realizes this approach is flawed due to air drag's influence. Instead, the correct method involves using the relationship v = dy/dt and understanding the conditions for a maximum in the function y(t). The integration of the velocity function is essential for determining the maximum height accurately.

PREREQUISITES
  • Understanding of basic physics concepts such as kinetic energy (KE) and potential energy (PE)
  • Familiarity with calculus, specifically integration and differentiation
  • Knowledge of projectile motion and the effects of air resistance
  • Ability to analyze functions and their maxima in mathematical terms
NEXT STEPS
  • Study the integration of velocity functions in calculus
  • Learn about the effects of air drag on projectile motion
  • Explore the conditions for finding maxima in functions, particularly in physics contexts
  • Investigate numerical methods for solving differential equations related to motion
USEFUL FOR

Students and professionals in physics, engineering, and mathematics who are interested in projectile motion analysis and the impact of air resistance on maximum height calculations.

sid9221
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http://dl.dropbox.com/u/33103477/gravity.png

I have worked out the first two bits. (So you can assume that I have that) But I can't figure out how to work out the maximum height reached.

I know there I can equal the KE and the PE to work out max height, but that doesn't look like it'll work here.
 
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Integrate v(t) to find y(t).
 
sid9221 said:
http://dl.dropbox.com/u/33103477/gravity.png

I have worked out the first two bits. (So you can assume that I have that) But I can't figure out how to work out the maximum height reached.

I know there I can equal the KE and the PE to work out max height, but that doesn't look like it'll work here.

The KE/PE argument will not work because of the presence of air drag. However, you don't need it. You have [itex]v = dy/dt.[/itex] Do you recall the conditions for a maximum of y(t)?

RGV
 
Last edited by a moderator:

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