A skier slides down a hill, starting with zero velocity at a height of 95 m above the bottom of the hill. The shape of the terrain is shown in Fig. 7-19



Figure 7-19. ( there is a picture, i don't know how to send it but the graph starts at 95 meters high then goes down, makes a curve to go up again, and another bump and it goes down again and stop. so basically one upward bump and downward bump and half bumps. if it does not make sense tell me. also the bumps are not 95 meters, they are shorter)

What is the velocity of the skier on top of the second, intermediate hill, whose height is 31 m?
I got 68.2 m/s, is that right? 1/2 mv^2= mg95 [/SIZE=7]


What is the skier's velocity at the bottom of the hill?
[SIZE=7](i know this one is 43.2 m/s)

Howru,
If one assumes zero friction, this turns out to be a simple conservation of energy problem.
The nice thing about conservation of energy is that it is path-independent. In other words, we don't need the figure!

What we know is the following:

V_initial= 0
H_initial= 95
H_final= 31

Now, from that we can calculate delta H (change in H) to be 64 meters.

Applying conservation of energy: mgh= (1/2)mv^2

solving for V we get V_final = sqrt(2g*deltaH)

and thus V_final = sqrt(2*9.81*64) = 35.4m/s

An interesting part of this answer is the fact that it doesnt involve mass, i.e. any object (when assuming these ideal conditions) will obtain the same final velocity when released along any path from one height to another.

[please correct if im wrong]

oh and you can check the other answer by plugging in delta H for the whole hill, i.e. 95m. (turns out you do have the right result.)