A car traveling at 22.0 m/s runs out of gas while traveling up a 25.0 degree slope. How far up the hill will it coast before starting to roll back down?
Vf^2 = V0^2 + 2ax
x = V0t + (1/2)at^2 (maybe?)
The Attempt at a Solution
I am using an online homework system called Mastering Physics. I did this problem and got an answer, which was marked wrong. Here's what I did:
Used Vf^2 = V0^2 + 2ax
where x = vertical position up the slope.
Vf = 0
V0 = 22sin25
a = -9.8
I solved for x, and got x=4.41 m. This is the vertical distance, or one leg of the right triangle across from 25 degrees. I used the law of sines to find the hypotenuse, and got 10.43 meters for the hypotenuse, which would be the distance that the car moved before it stopped.
This was marked wrong. Please explain why my thinking is wrong, and the right way to do the problem!! Thank you!!