How High Will the Toboggan Go on a Frictionless Hill?

[stylez]

Homework Statement

At the base of a frictionless icy hill that rises at 28.0 degrees above the horizontal, a toboggan has a speed of 12.0m/s toward the hill. How high vertically above the base will it go before stopping?

Homework Equations

height = [(v*sin (angle))^2] / (2g)

The Attempt at a Solution

height = [(v*sin (angle))^2] / (2g)
height = [(12m/s*sin(28degrees))^2] / (2(9.8m/s^2))
height = 31.74m^2/s^2 / 19.6m/s^2
height = approx 1.62m

I think I made a mistake with the velocity because this answer is incorrect. Can someone help me please and thank you.

 

[Dick]

I don't see any problem with that. I get the same thing.

 

[Anadyne]

I would do this problem with a more general equation like the conservation of energy (KE + PE = Etotal). It looks like the equation you gave was something that you didn't derive yourself and was specific to another situation. If you did derive this equation, rethink the velocity.

 

[Anadyne]

No, I disagree with his current answer. Dick, when using the conservation of energy, it doesn't matter which direction v is in. So there's no need to find a specific component of v.

 

[stylez]

In that case...

height = (v^2) / (2g)
height = (12m/s)^2 / (2*9.8m/s^2)
height = 144m^2/s^2 / 19.6m/s^2
height = 7.35m

or if I used energy...

KE=GPE
0.5mv^2 = mgh
0.5v^2 = gh (mass drops out)
0.5*(12m/s)^2 = h*9.8m/s^2
72m^2/s^2 = h*9.8m/s^2
(72m^2/s^2) / (9.8m/s^2) = 7.35m

This answer worked. Thanks Anadyne. Can't get em all Dick ;p

 

[Dick]

No, I disagree with his current answer. Dick, when using the conservation of energy, it doesn't matter which direction v is in. So there's no need to find a specific component of v.

Ooops. Sorry. Guess it's time to lay off the problem solving for tonite. Thanks for the correction.