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Projectile Flight (LAunch) |
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| May22-07, 07:16 PM | #1 |
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Projectile Flight (LAunch)
1. The problem statement, all variables and given/known data
Vi= 26.06 m/s Mass= 0.5 Kg Angle of launch= 45 degrees time of flight= 3.61 sec range= 64m I need to find the force given to the projectile on the cannon during the launch, and due to the formula F=MA, I am missing A, but I have not formula to find the acceleration of launch. 2. Relevant equations Which formula can be used to find the acceleartion of launch, and how to use it, for example calculate it with Vel in y or in x, 3. The attempt at a solution I have tried using Vf=Vi + AT Vf2=Vo2 + 2ax X=Vit + ½at^2 but each one throws different results, im not sure which one is the right to use, any help will b appreaciated. |
| May22-07, 07:20 PM | #2 |
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The projectile accelerates from 0 to 26.06m/s inside the barrel of the cannon. Do you have the barrel length? If you know the length of the cannon barrel you could use one of the formulas you gave to find a.
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| May22-07, 07:28 PM | #3 |
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yes the distance it displaces inside the cannon is of .47 meters
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| May22-07, 07:31 PM | #4 |
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Projectile Flight (LAunch)
Ok, you should be able to use one of the formulas you listed to find A. Now, which one involves the information you know and the variable your looking for?
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| May22-07, 07:36 PM | #5 |
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I guess that this one
Vf2=Vo2 + 2ax Vi is0 25.06^2=a 2 x .47 which gives me like 670m/s2 which is not accurate i think |
| May22-07, 07:38 PM | #6 |
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That is what I get as well. It looks right for the numbers you supplied.
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| May22-07, 07:40 PM | #7 |
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or maybe with the info i have i can find KE= 157 J
then W=FX 157/.47= F will that b accuratE? which is the same which i got with the previous information, thank you but the question, do those number look possible??? it was a projectile launched with ethanol... |
| May22-07, 07:42 PM | #8 |
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Yes, you could do it either way and should get the same answer if you calculated both correctly.
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| May22-07, 07:43 PM | #9 |
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Yep im calculateing them right, but hte question if that is those numbers actually look possible for a real life thing?
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| May22-07, 07:44 PM | #10 |
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157 J of Kinetic Energy seems plausible. What value do you get for the force?
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| May22-07, 07:55 PM | #11 |
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334 Newtons,
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| May22-07, 08:05 PM | #12 |
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It seems reasonable.
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| May22-07, 08:12 PM | #13 |
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thank you very much for ur help,
if by eany reason u know how can i calculate the enregy realsed by the combustion of ethanol.. i would appraeciated it :P |
| May22-07, 10:32 PM | #14 |
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Sorry, can't help you here, but I'm sure the answer can be found tabulated somewhere. |
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