Equations of speed and position under a constant force

[Raikou Tatsu]

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?

 

[quasar987]

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

 

[bomba923]

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 [itex] y\left( t \right) = 0s [/tex] to find your jump duration (*Note: $t \ne 0s$ )

The answer is 0.41 seconds

 

[lightgrav]

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.