How far up the hill will it coast before starting to roll back down?

[spartan55]

Homework Statement

A car traveling at 22.0 m/s runs out of gas while traveling up a 24.0deg slope. How far up the hill will it coast before starting to roll back down?

Homework Equations

(v_final)^2 = (v_initial)^2 + 2*(a_parallel)*(x_final - x_initial)
...and some trig

The Attempt at a Solution

I know the acceleration that is important is the acceleration parallel to the plane which is found by some trig manipulation:
(a_y)sin(24.0deg) = a_parallel, where (a_y) = -9.8 m/s^2.
Thus: a_parallel = -3.986 m/s^2.
Now would I use (v_initial) = 22.0 m/s and plug this with a_parallel, x_initial = 0, and v_final = 0 m/s into the above kinematic equation to get: 60.712 m ?
For some reason this just seems like too big of an answer. Any help appreciated. Thanks.

 

[Delphi51]

That answer looks good to me!
It is interesting to check it with an energy approach. KE at the bottom equals PE at the top so
mgh = .5*mv²
Solve for h, then convert that to a distance along the slope using trigonometry.

 

[spartan55]

Wow it is a lot easier with the energy approach. We haven't talked about that yet, we're still on basic kinematics. Thanks for your help!

 

[Delphi51]

Most welcome!