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).
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.
6 answers
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)
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)
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
So the anwser would be 784h = 1800
h=2.2
Is that correct?
I did the math wrong, I got 103 now.
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
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
1 answer