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Gavin

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Sep 27, 2009
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Have a question? Post it here! Have an answer to a question that alot of people ask? Post that here as well! This thread is for any questions, and answers, that you may about Busch Gardens Williamsburg.
 

Thomas

Never Sane to Begin With
Advisory Panel
Oct 24, 2009
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RE: Quick Questions and Answers

If I were to drop a rock which weighed 1.86 lbs off the top of Griffon, how long would it take to knock out a small child who disobeyed the safety warnings and was messing about in the flower bed below the drop?
 
Jul 28, 2011
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RE: Quick Questions and Answers

Alpenghöst said:
If I were to drop a rock which weighed 1.86 lbs off the top of Griffon, how long would it take to knock out a small child who disobeyed the safety warnings and was messing about in the flower bed below the drop?

Sounds like someone didn't do their physics homework! JK... Lol
 
Dec 23, 2011
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RE: Quick Questions and Answers

Alpenghöst said:
If I were to drop a rock which weighed 1.86 lbs off the top of Griffon, how long would it take to knock out a small child who disobeyed the safety warnings and was messing about in the flower bed below the drop?

I learned something like this in January!

I learned to place the situation on a graph, with an X-axis and a Y-axis. Which means the rock wouldn't go straight down, but veer off some. With all the information from the problem and that I put into the equations I used, this is what I did.

I first searched for the time to find an accurate velocity. To do this I used this equation (Final Distance = Initial Distance + Initial Velocity * Time + .5 * Acceleration * Time squared) with my Y-axis. I took out the Initial Distance since that was 0 and I took out the Initial Velocity * Time since the Initial Velocity is 0 and anything times 0 is 0. This leaves me with Final Distance = .5 * Acceleration * Time squared. I rearranged the problem to equal my Time and got Time = the square root of 2 * Final Distance divided by the Acceleration. Inserting numbers I get Time = 2 * 2m divided by a positive 9.8 m/s squared. I used 2m to suggest the rock may have landed 2m away from its original spot instead of right under where it was dropped. I ended up with a time of .64 seconds. In other words, it took .64 seconds for the rock to fall.

Now I used my new found time to get me my velocity. I used the equation the Velocity = Distance divided by Time. I used this equation on my X-axis portion of the problem. Time is the same for the Y and X axis's. So With that said, I got 62m divided by .64 seconds and got a Velocity of 96.88 m/s.

Of course this is how I have been taught in a public school setting and I had to change some numbers and add some numbers. I will ask my Physics teacher if I did the math right when I see him.
 

netdvn

Team Instinct
Jan 12, 2012
1,870
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in the rear
dvn225.deviantart.com
Weren't the starstruck KD fans the same ones who whined about all the changes 305 went through? :p

Oh and anyone planning on scoring one of those awesome plush chili peppers? I totally wanna add one of those to my growing anthropomorphic fruit collection.
 
Jul 22, 2010
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RE: Quick Questions and Answers

Party Rocker said:
I learned something like this in January!

I learned to place the situation on a graph, with an X-axis and a Y-axis. Which means the rock wouldn't go straight down, but veer off some. With all the information from the problem and that I put into the equations I used, this is what I did.

I first searched for the time to find an accurate velocity. To do this I used this equation (Final Distance = Initial Distance + Initial Velocity * Time + .5 * Acceleration * Time squared) with my Y-axis. I took out the Initial Distance since that was 0 and I took out the Initial Velocity * Time since the Initial Velocity is 0 and anything times 0 is 0. This leaves me with Final Distance = .5 * Acceleration * Time squared. I rearranged the problem to equal my Time and got Time = the square root of 2 * Final Distance divided by the Acceleration. Inserting numbers I get Time = 2 * 2m divided by a positive 9.8 m/s squared. I used 2m to suggest the rock may have landed 2m away from its original spot instead of right under where it was dropped. I ended up with a time of .64 seconds. In other words, it took .64 seconds for the rock to fall.

Now I used my new found time to get me my velocity. I used the equation the Velocity = Distance divided by Time. I used this equation on my X-axis portion of the problem. Time is the same for the Y and X axis's. So With that said, I got 62m divided by .64 seconds and got a Velocity of 96.88 m/s.

Of course this is how I have been taught in a public school setting and I had to change some numbers and add some numbers. I will ask my Physics teacher if I did the math right when I see him.

One of your parameters is very far off. I MIGHT take the time in a bit to figure out where you are off but not right now. Accurate time to impact is roughly 3.56 seconds at 512.63 joules or 77.7 mph. The fall height is not enough to quite terminal velocity.

::EDIT:: Your time is equivalent to ~1.9 meters ::EDIT::



Whoops, sorry, Swifty. :D
 
Apr 21, 2010
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Swiftman said:
Not even 24 hours after this thread opened and it has already gone up in flames. You know, I almost find this impressive.

Anyway, can we try to keep the questions serious?

Agreed.

This is the reason I created the Pointless Discussions thread.
 
Dec 23, 2011
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Yes, Atlantis, I know something had to be off, I was just working with what I had plus what I learned, currently, I am failing Physics, lol, go figure. However, if the questions were more like this one, I might be interested to actually not sleep in class, lol.

Here are some good questions:

How many steps do you take going down to the Rhine River on the Scotland side?
How many lights are in the park?
What exactly happens during the off season? Do they turn all the lights off and just leave or do they keep the lights on like they usually do when hey close?
How much is the park actually worth?
Do you think SWPE would sell the park individually in the near time future?
How many rides exactly are there?

Those are actually somewhat serious, I would like to know the answers to those one day.
 

Zachary

Administrator
Sep 23, 2009
15,831
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Virginia
Ok, this thread has two purposes:

1) Ask questions about the park.
2) Answer the questions of others.

This thread was created to eliminate those threads that consist of one question and then are pointless after an answer is received. This thread is NOT a place to ask questions just because you can. Have a purpose of some kind behind your question. How many lights are in the park is not a valid question. No one will know the answer to that.
 
Jul 13, 2011
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Orlando, FL
redsoxfan787 said:
Well, it's going to be an extra weekend longer :)
Haberdasher1973 said:
According to the Fun Tracker, it begins September 14th and runs through October 28th.

What were the start/end dates of previous HOSs? I'd say that (roughly) Sept 15 through Halloween has been the average...

EDIT: I didn't want to get the HOS 2012 Rumors thread off-topic with questions about prior years...
 

Zachary

Administrator
Sep 23, 2009
15,831
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Virginia
2010's ran from the 24th of September to the 31st of October (6 weeks)
2011's started on the 23rd of September and ended on the 30th of October (6 weeks)
 
Jul 13, 2011
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Orlando, FL
Swiftman said:
2010's ran from the 24th of September to the 31st of October (6 weeks)
2011's started on the 23rd of September and ended on the 30th of October (6 weeks)

Okay - I know that prior to those, though, I went to at least one opening weekend in mid September...

EDIT: perhaps it's just senility...
 
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