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http://img710.imageshack.us/img710/914/engkin.jpg!

Here is my kinematics argument. For now I am only going to look at ball 2 and ball 3. Make note of the following data.

$|v_0| = 10m/s$, $y_0 = 10m$, $\theta_2^0 = 30^0$, $\theta_3^0 = -45^0, g = -10m/s^2$

So that we have two equations

$y_2 = -5t^2 + |10|\sin30t + 10$

$y_3 = -5t^2 + |10|\sin(-45)t + 10$

Solving when they will hit the ground, I get $t_2 = 2s$ and $t_3 = 0.874s$

$y'_2 = -10t + |5|$

$y'_3 = -10t + 5\sqrt{2}$

solving I get $y'_2(2) = -15m/s$ and $y'_3(0.874s) = -15.81m/s$

EDIT: okay I was wrong, kinematics also gives me the correct answer (tested on my paper). Still intuitive to me.

They are different. Why is there a contradiction? It actually agrees with my original intuition. Ball 2 reaches a higher peak and because of the longer time it takes to come back, the velocity gained will be greater. Ball 3 just comes straight down.

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  • $\begingroup$ To make it intuitive that the speeds are the same, treat the vertical and horizontal components separately, and note that if you throw something up with a certain speed, when it comes down to its original height, it is going down with the same speed, so that the up and down ball are the same. For the horizontally thrown ball, you just need to know that the square of the vertical component is proportional to the height, but this is conservation of energy, and there is no better way to argue the equality. $\endgroup$
    – Ron Maimon
    Commented Dec 31, 2011 at 12:54

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You've forgotten to include the x' (i.e. horizontal velocity) components in your analysis. The energy equation correctly predicts that all three balls will have the same speed (which is a scalar) when they hit the ground; this is not to say, however, that they will have the same x and y components of velocity.

Try it out and see what I mean. Hope this helps!

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  • $\begingroup$ It is the same! But it's still very unintuitive to me. $\endgroup$
    – Lemon
    Commented Dec 31, 2011 at 0:18
  • $\begingroup$ Try solving the problem in general (i.e. only using symbols/variables) rather than plugging in arbitrarily chosen numbers; some of the cancellations/identities that lead to this result may be enlightening. $\endgroup$
    – Steven D.
    Commented Dec 31, 2011 at 0:20
  • $\begingroup$ yes I have. I know what you mean. I do that in math a lot, but I just needed a numerical answer to give me some real life thought. I am still in disbelieve. It just doesn't feel right to me. $\endgroup$
    – Lemon
    Commented Dec 31, 2011 at 0:25
  • $\begingroup$ As for intuition...all I can say is that it probably seems a little odd because, were you actually lobbing balls off a rooftop like this in real life, your results would be distorted by air resistance. Rest assured, though, that experiment would match theory much more satisfyingly if you conducted this experiment in a vacuum. :-) $\endgroup$
    – Steven D.
    Commented Dec 31, 2011 at 0:31
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    $\begingroup$ And for one more shameless plug: please feel free to "accept" my answer if it's been helpful. I'm new to the site and trying to build up some reputation! $\endgroup$
    – Steven D.
    Commented Dec 31, 2011 at 0:31

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