The conservation of energy tells us that, in the absence of friction, the total amount of energy in the system will remain constant. Since the skier begins each trip at the same height, H, and ends at the same height (the ground plane), the amount of energy in these systems is always the same.

PE(initial) + KE(initial) = PE(final) + KE(final)

The KE(initial) on each hill is 0. The PE(final) on each ground plane is 0.

So:

PE(initial) + 0 = 0 + KE(final)

PE(initial) = KE(final)

PE = mgh and KE = 1/2mv^2

mgh = 1/2mv2

gh = .5v^2

**√19.6h = v for each Point.**

If we include friction, we know that the final speed will be reduced, because friction is working against the forward motion created by gravity and the angle of the hill. The problem, as described in the introduction to this question, is that the angle of the hill changes, altering the normal force at numerous points, and thereby altering the portion of gravity that contributes to friction via the normal force.

However, this is not as difficult as it seems. The most direct path from the top of the hill to the Points on the ground would be a straight line. If we know the force of friction, we could calculate its action over this distance via W=Fd, thereby finding the energy that friction takes away from the skier.

While it would intuitively appear that a steeper hill would reduce friction,
and result in a faster skier at the end, this is not the case. True, the
steeper the hill, the less normal force, and the less friction, but a steep
hill results in a greater *shallow-*angled distance to cover to reach
one of the points, and shallow angles have a large normal force. **Thus,
any modification of the skier's path will be balanced by other parts of the
path, resulting in the same velocity at each point.**

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