Does Velocity Remain Constant in Vertical Motion?

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Discussion Overview

The discussion revolves around the concept of velocity in vertical motion, specifically addressing a scenario where a person bounces on a trampoline and returns to their initial height. Participants explore the relationship between initial and final speeds in the context of gravitational forces and energy conservation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant seeks clarification on why the speed upon returning to the initial height is the same as the initial speed of 4.0 m/s.
  • Another participant cites conservation of mechanical energy, stating that kinetic energy converts to potential energy at the peak and back to kinetic energy when returning.
  • A participant draws a parallel to throwing a rock straight up, suggesting that the speed upon return would also be the same, assuming no air resistance.
  • One participant notes that real-world factors like friction and air resistance would affect the actual speed upon return, emphasizing the distinction between speed and velocity.
  • Another participant explains that gravity is a conservative force, leading to the conclusion that kinetic energy at the initial height equals potential energy at the maximum height.
  • A later reply introduces Newton's laws to analyze the motion, providing equations for initial and final velocities, but does not resolve the discussion.

Areas of Agreement / Disagreement

Participants express differing views on the effects of real-world factors like air resistance, while some agree on the theoretical aspects of energy conservation and the behavior of velocity in ideal conditions. The discussion remains unresolved regarding the impact of external forces on the final speed.

Contextual Notes

Participants acknowledge that the analysis assumes ideal conditions without air resistance or friction, which may not hold true in practical scenarios.

revernance
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A need some help understanding the concept:

A person on a trampoline bounces straight upward with an initial speed of 4.0 m/s.

What is the person's speed when she returns to her initial height?

The answer is 4.0 m/s.

Can you guys explain why though?
Thanks so much!
 
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Conservation of mechanical energy.
At the top, all kinetic energy had been converted into potential energy; returning, that potential energy re-converted into kinetic energy.
 
What do you think is going on?

Hint: It's got nothing to do with trampolines. If you threw a rock straight up into the air with a speed of 4 m/s, what speed will it have when it falls back into your hand? (Ignoring air resistance, of course.)
 
Maybe what's confusing you is you're thinking of friction. The equation which produces that result ignores friction (which includes wind resistance, etc). So in the real world, no, you don't land with quite as much speed.

Also don't forget the difference between velocity and speed. Initial speed = final speed, sure. But initial velocity, +4.0 m/s in the y-axis (for trampoline example) yields -4.0m/s in the y-axis for final velocity.
 
gravity is a conservative force/./

when u throw the body and it reaches the max height, its velocity is 0( so KE is 0), whereas it has PE( which is = to mgh)

when it reaches back to the point of projection, its KE will be equal to the PE at highest point( since gravity is conservative force). therefore , the velocity will be equal to the velocity with which u throw...
 
or using Newtons laws..

initial velocity = u

at highest point v=0;

v=u+at;
a=g=10 m/s^2
u=10t;

now from highest point.

u=0;
v=0-10t(- ve for direction);
v= -10t;
 

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