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Studying formulas about velocity and acceleration I came up with a question: if I throw an object in the air with a velocity $v_0$ (suppose i throw it vertically) in how much time its final velocity $v_f$ will reduce to $0$ due to the force go gravity? Here is how I tried to solve the problem:

Calculation of the time

I know that the final velocity of a object that receive an acceleration is: $$v_f=v_0+at$$ where $a$ is the acceleration and $t$ is the time in which the acceleration acts. I supposed that $v_f$ after a negative acceleration (the gravitational acceleration on Earth $g$) will reduce to $0$ and so I set up the following equation: $$0=\vec{v_0}-\vec{g}\cdot t$$ and solving the equation for $t$ I got that \begin{equation} t=\frac{v_0}{g}\tag{1} \end{equation}

Calculation of the space

I know that the formula to calculate the space that is made by an object moving with an acceleration is $$S=v_0t+\frac12 at^2$$ But now I can apply $(1)$ to the equation: $$S=v_0\cdot \frac{v_0}{g}-\frac12 g\left(\frac{v_0}{g}\right)^2$$ $$S=\frac{v_0^2}{g}-\frac{v_0^2}{2g}=\frac{v_0^2}{2g}\tag{2}$$ That would be the formula for the space.


Reassuming an object thrown in the air with a velocity $v$ will stop moving in the air after a time $t=\frac{v}{g}$ after making a distance $S=\frac{v^2}{2g}$.


Is this correct?

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closed as off-topic by ja72, Brandon Enright, Kyle Kanos, David Z May 22 '14 at 21:16

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  • $\begingroup$ The time is $t=\frac{v_0}{g}$ and not $t=\frac{v^2}{g}$ you wrote in the end. Also your factor $0.051$ has units that you need to include to make the equation correct. $\endgroup$ – ja72 May 22 '14 at 20:12
  • $\begingroup$ I corrected the equations. Sorry for the error $\endgroup$ – Peterix May 22 '14 at 20:31
  • $\begingroup$ Why this question got -1? $\endgroup$ – Peterix May 22 '14 at 20:36
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    $\begingroup$ It is offtopic. Homework questions are to ask about specific concepts, not to check your results. Here is the policy: meta.physics.stackexchange.com/questions/714/… $\endgroup$ – Davidmh May 22 '14 at 21:00
  • $\begingroup$ This is NOT a home work question this is a problem that I invented BY MY OWN $\endgroup$ – Peterix May 23 '14 at 7:07
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Yes, that's how physics is done!

Aside from what I assume is a typo in your final summary, your equations (1) and (2) are both correct. You should note, however, that this is the Newtonian Way of answering your questions. Real-life experiments will show some variation in time and distance traveled, a quicker slow-down time, and a shorter path. This is due to air resistance.

You'll need a more complex model if you want super-accurate answers, but these should work for rough estimations and low-level physics classes.

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  • $\begingroup$ Thanks a lot! Also do you know if the object start instantaneously to fall or if it "stops" in the air, and if it does how much time does it take to start to fall down? $\endgroup$ – Peterix May 22 '14 at 20:35
  • $\begingroup$ The object will behave so that it's acceleration will always be constant. It constantly decelerates then constantly accelerates at the very same rate. This is also true when it's "stopping" mid-air. You could say that the time interval when the obejct is "very slow" is "very small", so if you requeire it to be stopped, then this time interval is "infinitely small" -> it is zero. The object stops just for an "infinitely small" time. $\endgroup$ – netom May 22 '14 at 21:08

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