QUESTION ABOUT ALTITUDE

[bogey1]

i dont know where to go for information and wouldn't know how to word it if i did a google search, so here goes. i have a thunder mug. its a piece of stainless steel (seamless) 6" tall 4" dia with a 2" bore. i packed some black powder in it and placed a (cro-cay) (dont know how to spell it) ball on top of it. the ball weighs about 1/4 lb. after ignition the ball took 18 seconds to reach earth. the mug was pointed straight up. im trying to find out how high the ball went. any ideas?

[Lycanthrope]

OK...I just drank a couple Lycanbrews and that may make me feel smarter than I am........

But......doesn't everything drop at an accleerated rate of 9.8M/s/s?

http://www.glenbrook.k12.il.us/GBSSC...Kin/U1L5b.html

I think we could figure it out from there. But no way could I do that until the morning.

[sjl127]

i dont know where to go for information and wouldn't know how to word it if i did a google search, so here goes. i have a thunder mug. its a piece of stainless steel (seamless) 6" tall 4" dia with a 2" bore. i packed some black powder in it and placed a (cro-cay) (dont know how to spell it) ball on top of it. the ball weighs about 1/4 lb. after ignition the ball took 18 seconds to reach earth. the mug was pointed straight up. im trying to find out how high the ball went. any ideas?

Are you kidding me? Do you know how much of a pain in the ass it is to do these calculations?

It's a lesson in physics and I would have to bust out my old college books about this stuff, but all I remember that would help you out now, is this...

The ball went about 868 yards high. Based on your 18 second observation.

Physics gurus, if you want to check me, assume 9 seconds up, and down based on similar deceleration and acceleration at 9.8 m/s2. Equation,

x=1/2 * a * t^2

[TaePo]

I'll get the easy part: croquet!

Swish.

[bogey1]

I'll get the easy part: croquet!

Swish.

i wasnt even close on that one........

[bogey1]

Are you kidding me? Do you know how much of a pain in the ass it is to do these calculations?

It's a lesson in physics and I would have to bust out my old college books about this stuff, but all I remember that would help you out now, is this...

The ball went about 868 yards high. Based on your 18 second observation.

Physics gurus, if you want to check me, assume 9 seconds up, and down based on similar deceleration and acceleration at 9.8 m/s2. Equation,

x=1/2 * a * t^2

dont ask me how i did it but i figured 2500 feet..

[Lycanthrope]

dont ask me how i did it but i figured 2500 feet..

Which is pretty close to 833 yards......near to SJ127's post.......

AT this point, I'd have said....damn high....near half a mile.

[bogey1]

Which is pretty close to 833 yards......near to SJ127's post.......

AT this point, I'd have said....damn high....near half a mile.

yep, thats what i was saying. sj127 must have done his homework....

[jb12string]

I am wondering if you can assume a ascent time of 9 second, If the ball can travel at greater speed on its trip up, it will be covering that half of the overall trip quicker than the descending half of the trip. I didn't read the formula, so I may be missing something but isn't it possible that the ascent took say, 6 seconds and the descent took 12 seconds?

[Philbert]

Are you kidding me? Do you know how much of a pain in the ass it is to do these calculations?

It's a lesson in physics and I would have to bust out my old college books about this stuff, but all I remember that would help you out now, is this...

The ball went about 868 yards high. Based on your 18 second observation.

Physics gurus, if you want to check me, assume 9 seconds up, and down based on similar deceleration and acceleration at 9.8 m/s2. Equation,

x=1/2 * a * t^2

That's OK as long as the ball did not reach terminal velocity, if it did then the height would be overestimated.