Kinematics and Work-Energy problem with skier

[alingy2]

Please look at attached picture.

Why did my teacher assume that vf does not change upon landing in the last sub-question? This makes no sense to me. Is there such a way that the skier can change direction of vf without changing the magnitude of it? Otherwise, the skier will collide in an inelastic collision with the ground (since he does not bounce off) and this means that all the vertical velocity is lost. I would calculate vertical velocity of vf and use that to find work done.

 

[BvU]

Excellent question. I think you are right in challenging teacher: after all most of the final speed is vertical (at least 36 m/s, horizontal at most 10 m/s) so landing parallel to the slope would require an angle of 75 degrees wrt horizontal, a lot more than the 35 degrees in the drawing.

(a 75 degree slope could arc to 0 degrees so that speed isn't lost -- clearly not provided for in the drawing).

The design of the ski ramp in the drawing is murderous. If you look at a real ramp you see that the ground follows the dotted line much closer than the straight one that says 100 m, so that skiers don't fall to their death.

 

[alingy2]

One doubt remains: even the skier lands parallel to the ground, there is still a collision, a loss of velocity, no?

 

[alingy2]

Can the skier ever land without losing some velocity?

 

[BvU]

If we factor out the friction, the speed lost is the speed perpendicular to the slope; you flex your knees to absorb that part of the kinetic energy. So the unlikely case where landing velocity and slope are parallel and the skiers trajectory is tangent to the slope at the touchdown point would not require flexing and not involve loss of tangential speed (which is all speed at that point). Subsequently following an arc to redirect the speed more horizontally doesn't necessarily require flexing -- it only requires some resistance to high g forces.