What is the equation of motion for two cubes colliding in vertical motion?

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The discussion revolves around the motion of two cubes thrown vertically, with the first cube launched with an initial velocity (v) and the second cube thrown five seconds later, also with the same velocity. The cubes collide at a height of 9 meters, prompting a calculation of their velocities at that point. Participants suggest using the equations of motion, specifically s = ut + 1/2 at^2 and v^2 = u^2 + 2as, to derive the necessary values. Vector form of the equation of motion is also recommended for a clearer understanding of the problem. The conversation emphasizes the importance of setting a proper coordinate system and understanding the physics involved in the collision.
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child play with cubes , he threw first one vertically to up with velocity of (v), after five seconds he threw second cube vertically to up with the same velocity , two cubes collided at 9 meter height what the velocity of two cube? (they are the same!)


actually i don't know but final answer, i need help please

and thank you all again , again , and again :-p .
 
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heaven eye said:
child play with cubes , he threw first one vertically to up with velocity of (v), after five seconds he threw second cube vertically to up with the same velocity , two cubes collided at 9 meter height what the velocity of two cube? (they are the same!)
You want to use s = ut + \frac{1}{2}at^2 and v^2 = u^2 + 2as setting s to 9 each time. Then the two equal each other and you can find a time as you know the acceleration and the time in each case.

The Bob (2004 ©)
 
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You should write the equation of motion in vector form

\vec{r}(t)=\vec{r}_{0}+\vec{v}_{0}t+\frac{1}{2}\vec{a}t^{2}

and then choose an axis along which the movement takes place,an origin for this axis, then a sense on this axis and then do the projection of the vector equality on this axis...

This is just the beginning,but it's essential.

Daniel.
 
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