Equations of speed and position under a constant force

AI Thread Summary
To determine how long it takes to stop rising after jumping upward with a speed of 2 m/s, apply the equations of motion under constant gravitational force. The key is to set the upward speed to zero, which indicates the moment of stopping. The correct formula to use is y(t) = t(2 m/s) - 0.5(9.8 m/s²)t², setting y(t) to zero to find the time. The calculated time for the upward motion is approximately 0.41 seconds. Understanding the wording of the question is crucial for accurately interpreting the conditions of the problem.
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 vertically,
*Set your initial position at y=0, then apply that equation
y\left( t \right) = t\left( {2\frac{m}{s}} \right) - \frac{{t^2 }}{2}\left( {9.8\frac{m}{{s^2 }}} \right).
Simply then, set y\left( t \right) = 0s [/tex] to find your jump duration (*Note: t \ne 0s <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:" />
 
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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