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Old Mar 31st 2009, 01:57 PM   #1
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speed explanation

Could someone explain to me why is the following case the two rocks have the same speed.
Two rocks are thrown from the top of a 50m cliff with the same speed, one up, the other down. When they hit the ground how is their speed?
I would think that the one thrown down would have a greater speed because of the force involved in throwing it down while the one thrown up had no force pushing it down besides its own mass and the gravity? I was told that they had the same speed when they hit the ground. I just does not make sense to me.

Thanks for any clarification.
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Old Mar 31st 2009, 03:35 PM   #2
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Indeed that is surprising but true. Are you familiar with the mechanical energy of a system? The formula is $\displaystyle E=\frac{mv^2}{2}+mgh$ and in your case it is conserved (because there is no dissipative force acting on any moment).
Note that the energy depends on the speed and not velocity and it also depends of the height.
Call $\displaystyle t=0$ when the rocks are being pitched.
At $\displaystyle t=0$, the energies of the rocks are equal. (Do you see why?).
And as the energy is conserved in this case, it means that they hit the ground at the same speed. (Look at the formula for the energy).
What is occurring is that if you throw a rock in the air (up direction), when it will pass at the same height it has been threw, it will already have a velocity. In fact the rocks reach its initial position with the same speed as it has been released but with an opposite direction. It explains it all.
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