Posted by **Joey** on Thursday, November 4, 2010 at 5:48pm.

If I have a roller coaster, and hill 1 height is 50cm and hill height 2 is 25 cm and the PE is .9 as it approaches the top of hill 2, how do I calculate the speed.

Please help

- Physics-Please help -
**bobpursley**, Thursday, November 4, 2010 at 5:52pm
THe PE is .9 where? Where did the coaster start?

- Physics-Please help -
**Joey**, Thursday, November 4, 2010 at 5:54pm
The coaster started at .18PE-the .9 is the PE at the top of the second hill.The second hill is lower.

- Physics-Please help -
**bobpursley**, Thursday, November 4, 2010 at 6:01pm
Well, the difference in the two Potential energies is now KEnergy, 1/2 mv^2, solve for v.

- Physics-Please check -
**Joey**, Thursday, November 4, 2010 at 6:21pm
. The way to calculate this is you take the KE would be .9. You would say .9 = ½ (.035 x v^2) = 2.215. Correct or no?

- Physics-Please check -
**Joey**, Thursday, November 4, 2010 at 7:31pm
Should I have done this as 1/2mv^2 = mgh

Cross out like terms and get to v^2 = square root of 2 x g x .25m(h)= 2.125 m/s

Would this be correct?

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