[hicheong]

The first rock is thrown horizontally with initial velocity v1, while the second rock is thrown at an angle of θ2 = 60 degree with initial velocity v2 = 2 v1. Rock 1 lands at a distance D1 form the base of the cliff, while rock 2 lands at a distance D2 form the base of the cliff.

D1 < D2
D1 = D2
D1 > D2

after I graph the problem, it looks like D2 will travel a greater distance than D1. However, I don't know how to prove it in a scientific way?

[ebaines]

We could go into a lot of algebra with formulas to prove that your conjecture that D2>D1 is correct. But there's an easy way to think this through:

For rock 2 its horizontal velocity component is 2 x v1 x cos(60) = v1. So bith rocks have the same magnitide of horizontal velocity, and consequently rock 2 will travel further than rock 1 if its time in the air is greater than rock 1's. When you throw a rock with an upward velocity component, after some period it will have reached the peak of its arc, and at that point its velocity is purely in the horizontal direction at velocity v1. Clearly the height of rock 2 at this point is higher than rock 1 was, and so it will take longer to fall to the ground from that point. So it has two advantages going for it: (1) it traveled horizontaly some distance before reaching its peak, and (2) it takes longer to fall to ground from its peak than rock 1 did. Both of these factor favor rock 2 going further than rock 1.