|Nov8-04, 08:59 PM||#1|
water balloon launching!
hey what does everyone know/estimate one of those three man giant elastic water ballon launchers has as its k constant in hooke's equation F = -kx
im trying to figure out a max range for one of those for a given strech length x and a given mass m.
i did 1/2 m v^2 = 1/2 k x^2 -> v = sqrt(kx^2/m) then put that all in the projectile formula basic stuff, all this seem sound? or do u think that the mass/projectile will not recive all of the spring energy? idk let me know what u guys think.
|Jun23-08, 03:04 PM||#2|
I am trying to work out the same thing with the goal of letting my students practice a little water balloon artillery. Try launching the balloon straight up. Calculate the total time of flight. take half the time and multiply that by 9.8 m/s/s that should be the speed at which the balloon hit the ground and the same speed that it left the launcher at. That should match your v from your PE = KE calculation. To veriffy. Launch the balloon horizontally from a given height. Calculate the time it will take for the balloon to fall from that height (d= .5(9.8m/s/s)t^2 gives you t = sqrt(d/2g).
Pull the balloon back the same distance you did on your vertical calculation and let the balloon fly horizontally. Multiply the time you get from calculating the vertical drop by your v from your other calculations and you should get where the balloon lands. If this works you can pulling to other lengths to see if you can get a value for k and then set up targets and work out the angle to hit them. Good luck let me know if this works.
|Jun23-08, 04:36 PM||#3|
These are made of latex rubber, so the "spring constant" isn't a constant.
Here is site with data for up to 300% strain (also used to launch radio control gliders).
For the heavy duty tubing, which is 24lbs at 300%, the tension is 27lbs at 350%. You can fill in the rest of the table with this info. I wouldn't go beyond 350%, because then a permanent stretch will result (or the tubing could break).
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