A car travelling at 30 m/s runs out of gas whiles travelling up a 20 degree slope. How far up the hill will it coast before staring to roll back down?
a = -9.8sin(20)
y=yo + Vo(t) + aT^2
Vf^2=Vi^2 + 2aY
The Attempt at a Solution
I tried finding the time the car rolls forward by using the second equation first:
0=30sin20 + (-9.8sin20)t
Then, I tried solving the vertical distance using either equation 1 or 3, both unsuccesfully.
s=(30sin20)(2.8) + (9.8sin20)(2.80^2)
giving: y= 55.007 y/sin20= 160.83
or yf^2=yo^2 + 2ax
0^2=30sin20^2 + 2(-9.8sin20)(x)
giving: y= 1.5506 or d= 4.475
Obviously, both of the attempted solutions are way off base of the answer, 132 m. I don't know where I went wrong, am I using the wrong equations or did I somehow plug in the numbers wrong?