Law of Conservation of Energy of a downhill skier

[darkmasterz8]

1. A skier is pushed from the top of a hill so that he starts moving down the hillside sloped at 27.6° to the horizontal with an initial speed of 0.434 m/s. After traveling 80.4 m, he reaches the bottom of the valley. Due to inertia, he then continues 70.4 m up another hillside sloped at 20.7° to the horizontal. What is the skier's speed when he reaches the top of the hill? Assume that you can neglect friction.



2. KE(b)+PE(b)=KE(a)+PE(a)



KE(b)+PE(b)=KE(a)+PE
1/2*m*(.434^2)+m*9.8*80.4=1/2*m*v^2+m*9.8*70.4---X out all m since they're the same both sides

.09+787.9=1/2*v^2+689.9
788=1/2*v^2+689.9
98.11=1/2*v^2
196=v^2
14=v

I already know the answer is 15.6 but I don't seem to get it through the work I've done. What I'm confused on is if 80.4 and 70.4 are the height? The problem makes it sound more like distance which doesn't make sense to me. I'm pretty sure that is the formula for law of conservation of energy.

 

[hy23]

yes 80.4 and 70.4 are the distances that the person travels...not the height, you will have to use the angle of inclination to find the actual height

and yes, you will encounter a lot of problems that give you distances to make your life harder

 

[darkmasterz8]

Well to use angle of inclination to find the height for initial, you do cosine 27.6=x/80.4 where x = 71.2 meters. The same thing with the after part. But when I plug in the height into the above equation, I still don't get it.

I'm not sure which of the three to use; cosine/tangent/sine because I don't know where the angle belongs. It says 27.6 to the horizontal so I assume the angle is at the top and you use cosine to find the adjacent given the hypotnuse of 80.4.

EDIT:
OOO I think I just got it. It's sine to get it.

Sine 27.6=x/80.4 where x=37.25
Sine 20.7=x/70.4 where x=24.77

Sub it in the above equation and I get 15.6 m/s.

Thanks for telling me about height/distance :)

 

[hy23]

draw a triangle with a right angle, the distance is the hypotenuse and the angle of inclination is opposite the height

this might be a good way to remember it ;)