How fast is the skier going just before she lands?

[ussjt]

An extreme skier, starting from rest, coasts down a mountain that makes an angle 25.0° with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts for a distance of 13.4 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3.20 m below the edge. How fast is she going just before she lands?

W= Fn*13.4

Fn*13.4= .5mv^2

(mg sin25 - (.200)(mg cos25))(13.4)= .5mv^2

(g sin25 - (.200)(g cos25))(13.4)= .5v^2

((9.8) sin25 - (.200)((9.8) cos25))(13.4)= .5v^2

31.6949624131 = .5v^2

2(31.6949624131) = v^2

sqrt (63.3899248261) = v

7.96177899882 = v
~~~~~~~~~~~~~~~~~~~~~~~~~~
Vy= 7.96177899882 (sin25)

Vy= 3.36479320085
~~~~~~~~~~~~~~~~~~~~~~~~~~

Vy = sqrt (3.36479320085^2 + 2(-9.8)(-3.2))

Vy = 8.604756 m/s

 

[Doc Al]

I didn't check your calculations, but it looks like you calculated the y-component of the final velocity. What happened to the x-component?

Tip: No need to find components of the velocity; you can use energy conservation to find the final speed.

 

[ussjt]

do I need the x? Because once it goes into free fall it is only accelerating in the y direction.

"you can use energy conservation to find the final speed"
I am confused by what you mean.

 

[Doc Al]

do I need the x? Because once it goes into free fall it is only accelerating in the y direction.

You are correct that only the y-component will accelerate, but the x-component of the velocity is needed to find the total velocity at the bottom. (It's just [itex]v^2 = v_x^2 + v_y^2[/itex].)

"you can use energy conservation to find the final speed"
I am confused by what you mean.

At the point that the skier leaves the cliff, she has a speed (which you figured out) and thus a kinetic energy. She also has gravitational PE compared to her landing point. Energy is conserved as she falls. (Gravitational PE is converted to KE as she falls.)