# Kinetic/Potential energy and thrown rocks

1. Oct 16, 2006

### bananan

A rock is thrown vertically upwards with a speed 'v' from the edge of a cliff. At the same moment, a second rock is thrown vertically downwards with the same initial speed 'v'. Which of the following statements regarding the motion of the rocks is true (ignore air resistance.)?

a. The rock which was thrown upwards reaches the bottom of the cliff with a higher velocity.
b. The rock which was thrown downwards reaches the bottom of the cliff with a higher velocity.
*c. Both rocks reach the bottom of the cliff with the same velocity at the same time.
d. Both rocks reach the bottom of the cliff with the same velocity but at different times.

The answer to this question is (C). Can someone please clearly explain to me WHY this is the case? How does the conservation of mechanical energy factor into the answer?

2. Oct 16, 2006

Think of the rocks as a system. What 'type' of energy does the system have before the rocks are thrown? Further on, what 'type' of energy does the system have just before the rocks reach the ground? (Assuming you define the 'zero' reference line of potential energy at the bottom of the cliff.)

3. Oct 16, 2006

### bananan

Oh, ok

Well, before they're thrown the rocks have only potential energy based on their height above the ground. Immediately prior to impact they have only kinetic energy.

Oh. Do they hit the ground at the same time because one system (the one in which the rock is thrown upwards) has extra potential energy injected into it, so to speak, and the other has an equivalent amount of extra kinetic energy added to it?

If this is correct reasoning, how would it be expressed mathematically?

4. Oct 16, 2006

Right, that's a way to look at it, since, at every two moments 1 and 2, $$E_{k1} + E_{p1} = E_{k2}+E_{p2}$$. (Which means $$E_{2}-E_{1} = \Delta E = 0$$, i.e. energy is conserved.)