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.
__________________ **Isaac** If the problem is too hard just let the Universe solve it. |