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?
Answer
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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