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I have this problem:

A student throws a set of keys vertically upward to her sorority sister, who is in a window 3.30 m above. The second student catches the keys 1.60 s later

1) With what initial velocity were the keys thrown?

2) What was the velocity of the keys just before they were caught?

For question #1 I got 10 m/s, which is correct. However, for the second question I also got the right answer but at first I did not know what formula to use since there are two kinematic equations specific for final Velocity

(1) $v = v_o + at $

(2) $v^2 = {v_o}^2 + 2ad $

I know the correct result comes from using formula (1) but I do not understand why using both formulas I get different results. How then would I know which formula to use in different situations?

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    $\begingroup$ Both are equivalent and should give the same result. If one doesn't, that means you made a mistake. $\endgroup$ – Javier Sep 8 '17 at 21:11
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I know the correct result comes from using formula (1) but I do not understand why using both formulas I get different results. How then would I know which formula to use in different situations?

I'm not sure why you're using formula (1) as it has two unknowns $v$ and $v_0$; the final and initial velocities respectively.

Assuming you took gravitational acceleration $a=-g\approx -10 \mathrm{ms^{-2}}$ then I would use $$d=v_0 t - \frac12 g t^2$$ which when solving for $v_0$ gives $$v_0 =\frac{3.3+0.5\cdot 10 \cdot 1.6^2}{1.6}=10.0625\approx 10\mathrm{ms^{-1}}$$ for the intitial velocity.

Then using formula (2) we have

$$v=\sqrt{10.0625^2-2\cdot 10 \cdot 3.3}=5.9375\mathrm{ms^{-1}}$$ for the final velocity.

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The difficulty is that the keys could have been caught before or after they reached the peak of their flight.

Using vf^2 = v0^2 + 2ad you can insert 0 as final velocity and -9.8 m/s^2 as a. Solving for d you will find the highest the keys could go is about 5.1 meters.

I'm guessing you used x = x0 + v0*t + 1/2at^2 to solve part 1.

What isn't clear from that solution is that the key was caught after the keys had reached the Apex. You can find this by using v = v0 + at and finding the time at which velocity is 0 (max height). This is around 1 second.

Now we get a clearer picture of the whole situation. They keys are thrown up and caught as they are falling down.

We know from this that the velocity of the keys is negative.

The first equation will give you this answer correctly as you noted.

The second equation had the velocity squared. We need to remember that when solving for any squared variable, there are actually two answers, the positive and negative square root.

The reason the second equation gives two results is that there are two times where the keys are at 3.3m. One is on the way up and one is on the way down. The velocity on the way up is the positive square root and the velocity on the way down is the negative square root.

Hope this helps!

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The two equations are the same, but each obtained by different methods.

The first one (1) is the velocity equation for a body in movement with constant acceleration:

$v = v_{0} + at$

Where $a$ is the gravitational acceleration.

The other one (2) is a derivation of energ conservation. In the first stage (when the student throws the keys), there is only kinetic energy, in the second one, there are kinetic energy and potential energy:

$\frac{1}{2}mv_{0}^{2}=\frac{1}{2}mv^{2}+mgd$

If we reordenate the equation, and knowing that $a=-g$, we obtain:

$v^{2}=v_{0}^{2}+2ad$

In summary, both equations must give you the same result.

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  • $\begingroup$ What you say is true, but it does not neccesarily come from energy conservation. (2) can be deduced by putting together the two equations of MRUA: $v_F=v_o+at$ and $\Delta s=v_o t+at^2/2$. $\endgroup$ – FGSUZ Sep 8 '17 at 22:15

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