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?
(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.