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March 1, 2015

Posted by **Hannah** on Wednesday, May 30, 2012 at 9:17pm.

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.

- Physics(Please respond) -
**Damon**, Wednesday, May 30, 2012 at 9:31pmm 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).

- correction -
**Damon**, Wednesday, May 30, 2012 at 9:34pmremember 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)

- Physics(Please respond) -
**Hannah**, Wednesday, May 30, 2012 at 9:46pmso 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

- Physics(Please respond) -
**Hannah**, Wednesday, May 30, 2012 at 9:46pmIs that correct?

- Physics(Please respond) -
**Hannah**, Wednesday, May 30, 2012 at 9:48pmI did the math wrong, I got 103 now.

- Physics(Please respond) -
**Elena**, Thursday, May 31, 2012 at 4:45pmIf 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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