# I think I disprove this with kinematics, but energy says it is right

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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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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. – Ron Maimon Dec 31 '11 at 12:54