#### Gavin

##### root@parkfans.net:/# rm -r *▌
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.

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?

• IndyRacingNut and Party Rocker
*Calls Atlantis*

• Haberdasher1973
From the very top to bottom, V=35m/s.

Remeber Force= MassxAcceleration.

1.86lbs converts to .844kg (rounded off for adequate significant figures of course!)

Which would equate to 29.5 newtons, AKA more than enough to teach an eternal lesson.

• K8theGr8, Owls R A Hoot, Reece and 4 others
What if you dropped a small child from the top of Griffon?

• kn0ebels
Shafor, what is the weight of this small child?

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

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.

So here's a question: How come the park can't do anything good enough for us?

Edit: And by "us", I mean "you guys" • Flyer and Merboy
Because, unlike starstuck KD Fans, we actually have brains and opinions.

Weren't the starstruck KD fans the same ones who whined about all the changes 305 went through? 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.

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?

• Haberdasher1973 and Pretzel Kaiser

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. 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.

• Haberdasher1973 and Zachary
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.

Ok, this thread has two purposes:

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.

• Haberdasher1973, FAL NR OX, Pretzel Kaiser and 1 other person
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...

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)

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...

How many miles is it around the main park path?

• Party Rocker
Consider Donating to Hide This Ad