Asked by Hannah

a rollar coaster is 80m high and has a 50kg car at rest. Ignore air resistance and friction.

If the speed of the car is 60m/s what is the height?

0+50(9.8)(80) = 1/2(50)(vf^2) + (60)^2 + 50(9.8)(h).

Is this equation correct?? If so I do not know how to solve for h since on the right side there is the h and vf^2.

Answers

Answered by Damon
m g * falling distance = (1/2)m (60)^2
m cancels
Find falling distance, subtract from 80
Your equation is right but note that 50 cancels and you really have m g (80-h).
Answered by Damon
remember that vf = 60 is given
If I wrote all this your way I would have
0 + m g (80) = (1/2)m(60^2) + m g (h)
or
9.8(80-h) = (1/2)(3600)
Answered by Hannah
so on the left side of the equation the mass is gone and on the right side the mass and gravity is gone.

So the anwser would be 784h = 1800

h=2.2
Answered by Hannah
Is that correct?
Answered by Hannah
I did the math wrong, I got 103 now.
Answered by Elena
If the roller-coaster starts from rest at the hill ita speed at the bottom of the 80- m hill is
m•g•h =m•v²/2,
v=sqrt(2•g•h) = sqrt(2•9.8•80)= 39.6 m/s.
Therefore, I believe that you have the mistake in your given data: the velocity may be
60 km/hr = 16.67 m/s. Then the problem makes sense.
So, if we have to find at what height the car has the velocity “v”,the solution is
mg(H-h) = mv²/2.
h = H - v²/2•g = 80 – (16.67) ²/2•9.8 = 65.8 m

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