Posted by Joey on .
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

PhysicsPlease help 
bobpursley,
THe PE is .9 where? Where did the coaster start?

PhysicsPlease help 
Joey,
The coaster started at .18PEthe .9 is the PE at the top of the second hill.The second hill is lower.

PhysicsPlease help 
bobpursley,
Well, the difference in the two Potential energies is now KEnergy, 1/2 mv^2, solve for v.

PhysicsPlease check 
Joey,
. 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?

PhysicsPlease check 
Joey,
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?