# Fun Physics problems with coasters

Saturday, January 13, 2001 5:53 PM
I have a physics problem for you all and it goes like this: you have a roller coaster with a height of 240 ft. Calculate the speed at the end of the drop if the mass of the train is 2000kg.
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Saturday, January 13, 2001 6:01 PM

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2000 stats: 135 coasters in 26 parks
Not Too Shabby For A Summer
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Saturday, January 13, 2001 6:05 PM
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Saturday, January 13, 2001 6:14 PM
What's the angle of the drop?
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Saturday, January 13, 2001 6:17 PM
in this problem, ignore angles, air resistance, and friction. It's crude to do that, but the problem can still be solved *** This post was edited by smoothncomfy on 1/13/2001. ***
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Saturday, January 13, 2001 7:10 PM
84-1/2 miles per hour. Mass and angle are unimportant in this case, so long as we're ignoring resistive forces...

--Dave Althoff, Jr.
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Saturday, January 13, 2001 7:21 PM
Mass has no effect
32 ft/s/s

Time Distance Speed average speed
1 s----16 ft----32 ft/s---16 ft/s
2 s----64 ft----64 ft/s---48 ft/s
3 s---144 ft----96 ft/s---80 ft/s
4 s---256 ft---128 ft/s--112 st/s

So at about 240 ft traveled the object would be traveling 124 ft/s

124 ft/s x 60 sec/min x 60 min/hr
--------------------------------------- = 84.5 MPH
5280 ft/mile

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#1 Steel-Incredible Hulk
#1 Wood--Timber Wolf *** This post was edited by T-Wolf on 1/14/2001. ***
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Saturday, January 13, 2001 7:23 PM
Get a life!!!!!!
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Saturday, January 13, 2001 7:26 PM
well, I always use the work-energy theory to solve these
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Saturday, January 13, 2001 7:27 PM

coasterbabez said:
Get a life!!!!!!"

I can't see how what other people do in their spare time affects you in any way...

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-Dave Rutherford
*If things seem under control, you're just not going fast enough*

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Saturday, January 13, 2001 7:46 PM
Coasterbabez:
Physics is a major part of my job and it lets me understand my hobbie alot more than people who have a life.
Before you can have any fun on a coaster there are some people who sit in front of a computer without a life who use physics to calculate the forces exerted on a rider and the limits it can obtain. So that after riding a coaster you literally have a life.

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#1 Steel-Incredible Hulk
#1 Wood--Timber Wolf
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Saturday, January 13, 2001 7:49 PM
Nicely stated, T-Wolf.
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Saturday, January 13, 2001 7:56 PM
Thanks, do you get the same answer as me and rideman?
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#1 Steel-Incredible Hulk
#1 Wood--Timber Wolf
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Saturday, January 13, 2001 7:58 PM
yeah, but I got 84.1 mph using the work-energy theory, but your shortcut works just as well I guess.
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Saturday, January 13, 2001 8:05 PM

"On The Eigth day they made Coasters"

Let me guess that would mean that "They" need to get a life?

I would have posted it there if it wouldn't have been closed.
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#1 Steel-Incredible Hulk
#1 Wood--Timber Wolf
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Saturday, January 13, 2001 9:04 PM
on my web page I will soon put in original drawings of lift hills and you can see my process on the work-energy theory.
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Sunday, January 14, 2001 11:40 AM
Well, here's the equation I used:
` V [mph] = 60/11 * sqr(h[feet]) `

As for where that comes from...
` A = V' = D'' = 32 ft/sec/sec. V = At + V0 = 32t ft/sec + 0 ft/sec = 32 ft/sec. D = 1/2At^2 + V0t + D0 = 1/2At^2+0t+0 feet = 16t^2 feet `
Now, since we know the distance, we can rearrange and substitute as follows:
` t = sqr(d)/4 seconds v = 32 sqr(d)/4 feet per second `
Given a 240 foot fall...
` t = sqr(240)/4 = 3.8729833462 seconds v = 32 sqr(240)/4 = 123.935467078 feet/second `
Multiply by 3600 seconds/5280 feet and get...
`84.5014548259 mph`

To generalize, let's go back to the velocity equation:
` v = 32 feet * sqr(h)/4 seconds * 3600 seconds/5280 feet v = (115200/21120) * sqr(h) miles per hour v = (60/11) * sqr (h) miles per hour `

--Dave Althoff, Jr.
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Sunday, January 14, 2001 12:00 PM
Give us another problem smoothncomfy :)

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As Antuan goes through school withdrawal...
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Sunday, January 14, 2001 5:40 PM

What is the Average Speed of the Millennium Force from point A, located at the top of the lift to point B, located at the first break?

I am willing to bet that the Force WASTE all other (complete circuit) coasters in this category.

To do this we need the distance between B and A and the time it takes the train to travel between A and B. I can talk with Pythagorus and get the length of the lift, but I don't know the length of the brake run and Station, nor the time it takes for the train to get from A to B.
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Sunday, January 14, 2001 5:44 PM
How about from the time the train leaves the station to the time it reaches the brake run, considering how the train is accelerated up the lift at 13 mph.
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