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Tower of Terror physics problem (1 Viewer)

Master Gopher

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Question on the physics involved with the Tower of Terror ride at Dreamworld (pics and some stats)

This isn't a HSC question, but nobody else in prelim can figure it out either...

"By combining the equations for kinetic energy and gravitational potential energy, calculate the expected height that the car will reach. You will need to consider the conservation of energy."

Formula for gravitational potential energy:

Ep = -G (m1 x m2 / r)

where G = 6.67 x 10^-11
m1 and m2 = the masses of the two objects
r = distance between their centres

NB: The car is accelerated over a short distance at the start of the ride, after which it is carried up the tower by its own momentum.

Data:
mass of car = 6000kg
max velocity = 161km/h (44.72m/s)
acceleration time = 3 secs
hence a at end of acceleration zone = 14.9m/s/s
Ek = 134,166.7 joules

I have no idea how to work this out! I've tried various things at random but it's not getting anywhere, has anyone done a question like this before?

Edit:// Cross-posted to Year 11
 

DAAVE

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Dude!

For the height of the tower, E = mgh will suffice.

At the bottom of the tower the E = 1/2 * m * v^2.

All of this KE is converted into GPE, so

GPE (at the top) = KE (at the bottom):

mgh = .5mv^2

h = (.5 * v^2) / g (the masses cancel)

I'll let you sub in the numbers. It's worthy to note that it's independant of the mass. Energy analysis is really, really useful in physics (really useful).

You could have done it with the other formula, but for such a relatively small height, mgh is more than accurate enough.
 

Master Gopher

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Thankyou very very much!!

After about 1/2 hour of fiddling around with the Law of gravitation formula, I knew there had to be a better way - besides it's way too inaccurate for distances that small.

Thanks again - I haven't actually been introduced to E=mgh before, it wasn't in any of my texts oddly enough.
:)
 

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