Asked 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.
Answered by Henry
h = (V^2-Vo^2)/2g.
h = (0-900)/-20 = 45 m.
Answered by Henry
Correctin:
Vo = 30m/s @ 30o.
Yo = 30*sin30 = 15 m/s.

h = (Y^2-Yo^2)/2g.
h = (0-225)/-20 = 11.25 m.
Answered by Anonymous
The answer is 90 meters. Henry don't know.
Answered by Anonymous
Yes. The answer is indeed 90 meters.
Answered by Anonymous
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.
Answered by Anonymous
all these answers are wrong
Answered by Anonymous
it 90 meters y'all don't know what you are talking about
Answered by Tarun Karthic
it be 90 meters yo. Getcho facts straight yo
Answered by Anonymous
Its actually 11.7 if youre on Quest from UT
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