posted by Anonymous .
A runaway truck lane heads uphill at 30◦ to the horizontal.
If an out of control 15000 kg truck enters the lane going at 30 m/s , how far along the ramp does it go? The acceleration due to gravity is 10 m/s2.
h = (V^2-Vo^2)/2g.
h = (0-900)/-20 = 45 m.
Vo = 30m/s @ 30o.
Yo = 30*sin30 = 15 m/s.
h = (Y^2-Yo^2)/2g.
h = (0-225)/-20 = 11.25 m.
The answer is 90 meters. Henry don't know.
Yes. The answer is indeed 90 meters.
Yes, the answer is 90 meters because h=(Vo^2-V^2)/g. g stands for gravity, which is 10 m/s^2. I don't know where Henry got the two from, but it doesn't go there. Anyways... h=(900-0)/10, which = 90 meters. Sorry Henry. Much luck to you in your future in physics.