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post #1 of (permalink) Old 01-10-2006, 09:04 AM Thread Starter
 
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O/T Water Rocket calculations


"I shot a ball into the air.
12 seconds later it landed there.
What I'd like to know, without splitting hairs
is how far the ball went into the air. "

OK so much for the poetic stuff. I was fooling around with a new water bottle launcher and just did a time estimate for height but it came out to some improble height like 576 feet.

It was a 2 liter bottle wit just air at around 65 psi, no water, with a raquet ball on top. The bottle dropped away but the ball kept on climbing.

So how far did it realy rise. What equasion did you use?
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post #2 of (permalink) Old 01-10-2006, 09:33 AM
 
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Re: O/T Water Rocket calculations

I can't help you on the calculation, although I'd like to know the formula also. However, what does this contraption look like? Anything like this?
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File Type: jpg 1475843-launch.JPG (89.2 KB, 23 views)
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post #3 of (permalink) Old 01-10-2006, 09:35 AM
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Re: O/T Water Rocket calculations

D= Di + Vi*t + (1/2)* a *(t^(2))

D is distance
Di is initial position...in this case is zero
Vi is initial velocity....also zero since we are starting at rest
a is acceleration from gravity (9.8m/s or 32.2 ft/s)
t is time.

unless we know the air resistance and all that we cant get a perfect answer.....so we'll have to pretend that the time to max altitude is the same as the time to hit the ground from max altitude. so the easiest way is to take your total air time, divide by 2, and use that in D=(1/2)(a)(t^2) to find the distance traveled in that time increment, which would be your altitude
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post #4 of (permalink) Old 01-10-2006, 09:54 AM
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Re: O/T Water Rocket calculations

12 seconds is a darn long time! Throw a rock up into the air and count how long it stays up there. Not 12 seconds.

The only error in the post is that the units for acceleration are ft/s^2, not ft/s.

So, using half the time to rise, and half to fall, we can find the distance fallen in 6 seconds due to gravity, starting from rest.

1/2 * 32.2 ft/sec^2 x (6 sec)^2

units for seconds-squared cancel out, leaving us units of feet.
=

579.6 feet.

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post #5 of (permalink) Old 01-10-2006, 09:59 AM
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Re: O/T Water Rocket calculations

yea thats what i meant....m/s^2 and ft/s^2...... whoops

it prolly wont be the actual height though. since you have an empty bottle coming down and a full bottle going up your going to have differences in the travel time going up vs going down. (this wouldnt happen if we lived in a vacuum...but we dont)

EDIT. just for example if the time was just one sec less (t=5 sec) we'd have only 400 feet. best bet would be to try to estimate the time from launch to peak instead of taking total time divided by 2
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post #6 of (permalink) Old 01-10-2006, 10:04 AM
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Re: O/T Water Rocket calculations

I can't tell you how far it went up, but I am very interested in how you made this rocket. Can you refer me to a website, or post some simple instructions. My dad teaches physics at Cal Poly and would also be interested.

TIA
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post #7 of (permalink) Old 01-10-2006, 10:04 AM
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Re: O/T Water Rocket calculations

Pete is right, if you ignore two things: Aerodynamic drag going up and aerodynamic drag coming down. The ball probably reaches terminal velocity on the way down - racquet balls aren't very dense. And if that's so it must have exceeded terminal velocity at launch. The ball almost certainly took more time to fall than to rise.

Two ways to get an accurate answer:

1 Find the coefficient of drag of the ball, the weight of the ball, and a knowledge of calculus, and do like the constipated mathematician - work it out with a pencil.

2 Time exactly how long the ball falls. Take it to a tall building and drop it out of windows on different floors until it takes that amount of time to hit the ground.

My guess is 300 feet - about half of the theoretical distance in a vacuum.
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post #8 of (permalink) Old 01-10-2006, 10:11 AM Thread Starter
 
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Re: O/T Water Rocket calculations

Maybe not live in a vacuum but I seem to work in a vacuum at times.

I'll take a picture of my laucher later but it's made out of CPCV shedule 40 pipe. The release is 4 wire ties held in place with a slip collar. Pull the collar down and off it shoots.

Doesn't 579 feet seem awful high though? The rocket never got as high as the ball did.

I think we are going to make an angle sighter to get a real measurement.
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post #9 of (permalink) Old 01-10-2006, 10:13 AM Thread Starter
 
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Re: O/T Water Rocket calculations

[ QUOTE ]
I can't tell you how far it went up, but I am very interested in how you made this rocket. Can you refer me to a website, or post some simple instructions. My dad teaches physics at Cal Poly and would also be interested.

TIA

[/ QUOTE ]

Here is a good place to start with simulators exc.
Water rockets

Just another reason for OBA. [img]/ubbthreads/images/graemlins/smile.gif[/img]
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post #10 of (permalink) Old 01-10-2006, 10:42 AM
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Re: O/T Water Rocket calculations

Yes, air resistance will cause the distance to be less, but 300ft (cutting it approx. in half) seems a low estimate to me - that's the length of a football field, standing on end, and we're still talking 12 seconds of air time, and I always thought racketballs were pretty dense -- small and kinda hard.

Just my $0.02.
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