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 $y'_3(0.874s) = -15.81m/s$

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.

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.