Energy and roller coaster question

[enapper]

A roller coaster starts at height (h). It goes down this hill and then goes up a second hill that is at a height of 28.5 m at a speed of 15.8 m/s. How high was the initial hill?


Given equations:
U=mgh
KE=1/2mv^2
W=Fd
P=W/T



Honestly, we've never done a problem like this before and I'm unsure how to solve it. I know potential energy is at it's highest point at the peak of the first hill and that energy is always conserved, so I said mgh(1)-1/2mv^2(1)=mgh(2)-1/2mv^2, but I didn't get a logical answer. Can anyone help me?

 

[tiny-tim]

welcome to pf!

hi enapper! welcome to pf!

(try using the X² and X₂ icons just above the Reply box )

your formula should work …

show us your full calculations, and then we'll see what went wrong

 

[enapper]

Thanks!

mgh=mgh(2)-1/2mv^2(2)

I canceled out the mass because it should be the same throughout the equation and no mass is given, leaving me with:
gh=gh(sub2)-1/2v^2(sub2)

(9.8)(h)=(9.8)(28.5)-(1/2)(15.8)^2

9.8h=279.3-124.82

h=154.48/9.8

h=15.76

This answer seems illogical, though, because normally the cart would not be able to exceed the initial height, and the second hill is much higher than 15.76m.

 

[tiny-tim]

oh, on second thoughts your formula was wrong …

it should have a + not a - …

conservation of energy is KE + PE = constant

 

[enapper]

That makes sense. "The whole is equal to the sum of the parts" sort of thing.
Thanks a million!