Fun Physics problems with coasters

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.

the answer is: C

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2000 stats: 135 coasters in 26 parks
Not Too Shabby For A Summer

is that your final answer?

What's the angle of the drop?

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. ***

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.

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. ***

Here's the answer:
Get a life!!!!!!

well, I always use the work-energy theory to solve these

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coasterbabez said:
"Here's the answer:
Get a life!!!!!!"

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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*

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

Nicely stated, T-Wolf.

Thanks, do you get the same answer as me and rideman?
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#1 Steel-Incredible Hulk
#1 Wood--Timber Wolf

yeah, but I got 84.1 mph using the work-energy theory, but your shortcut works just as well I guess.

In another thread that has already been closed coasterbabez said:

"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

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.

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.

Give us another problem smoothncomfy :)

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As Antuan goes through school withdrawal...

How about a tough Question.

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.

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.