Equations of speed and position under a constant force

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Homework Help Overview

The discussion revolves around a physics problem involving the motion of an object under the influence of gravity, specifically analyzing the time it takes for a person to stop rising after jumping upward with an initial speed of 2 m/s.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster seeks hints on how to approach the problem, while some participants suggest using equations of motion under constant acceleration due to gravity. Others discuss the implications of air resistance and the interpretation of the question's wording.

Discussion Status

Participants are exploring different interpretations of the problem and discussing various equations that could be applied. Some guidance has been offered regarding the use of specific equations and the importance of understanding the conditions under which the timing stops.

Contextual Notes

There is an assumption of no air resistance in the problem, and participants are questioning how to accurately interpret the wording of the question to determine when to stop timing the upward motion.

Raikou Tatsu
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there's a question in my book that says "If you jump upward with a speed of 2 m/s, how long will it take before you stop rising?" anyone have a hint as to how i would go about answering this?
 
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Use the equations of speed and position under a constant force (in this case the gravitational force).

v(t) = v_0 + a*t
x(t) = x_0 + v_0*t + 0.5at²
 
Raikou Tatsu said:
there's a question in my book that says "If you jump upward with a speed of 2 m/s, how long will it take before you stop rising?" anyone have a hint as to how i would go about answering this?

Assuming no air resistance, right?
Since you're jumping [itex]vertically[/itex],
*Set your initial position at y=0, then apply that equation
[tex]y\left( t \right) = t\left( {2\frac{m}{s}} \right) - \frac{{t^2 }}{2}\left( {9.8\frac{m}{{s^2 }}} \right)[/tex].
Simply then, set [itex]y\left( t \right) = 0s [/tex] to find your jump duration (*Note: [itex]t \ne 0s[/itex] <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" /> )<br /> <br /> The answer is 0.41 seconds [/color] <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f600.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":biggrin:" title="Big Grin :biggrin:" data-smilie="8"data-shortname=":biggrin:" />[/itex]
 
1) Bomba's "jump duration" is 2x as long as the
duration of upward travel. No big deal ...

BUT:

2) It is important to find out how to READ the WORDS of a question!
Otherwise it's going to be a long, hard, confusing, frustrating year.
The key is knowing what event-condition tells you to stop timing...
here, "stop rising" is translated into "upward speed = 0".

So Quasar's first equation is all you need to answer this question.
Bomba's approach will get you the right answer
(if you divide by 2, and if there's no air resistance)
but can't be generalized to, say, when does a police car catch up.
Quasar's APPROACH even works (slight mod of eq'n) if there IS drag.
 

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