Speed up a hill

  • Thread starter bertoline
  • Start date
  • #1
8
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Homework Statement


At the base of a frictionless icy hill that rises at 29.0deg above the horizontal, a toboggan has a speed of 12.0 m/s toward the hill. How high vertically above the base will it go before stopping?


Homework Equations


Wtot=K2-K1
K=1/2mv^2
K2=1/2mv^2=K1+Wtot


The Attempt at a Solution


Vx=12cos(29) = 10.5 m/s
vf^2-vi^2=2as
0-(10.5)^2=2(-9.8)s
s= 5.625
 

Answers and Replies

  • #2
130
0
This is how i would do it:
1. draw an fbd of the toboggan
2. use f=ma to solve for a
3. solve for distance travelled on the hill
4. then you make a triangle, with the distance you found in 3) being the hypotenuse, solve for the vertical side, and thats how high it has travelled.
 

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