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Ball Launched from spring gun (max height) 
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#1
Apr2410, 05:54 PM

P: 46

1. The problem statement, all variables and given/known data
A ball of mass m is launched at an angle theta from a spring gun of length D. The lower end of the gun, where the spring is attatched, is at ground level. The spring has spring constant k and an unstretched length D. Before launch, the spring is compressed to a length d. Determine the maximum height h above the ground the ball reaches. 2. Relevant equations (k(x^2)/2)=mgh ?? x = mgcos(theta) 3. The attempt at a solution Plugging in "x" I get: h=(k(mgcos(theta))^2)/2mg Is this correct? If not please explain in detail and provide the steps and solution. TIA. 


#2
Apr2410, 06:08 PM

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P: 41,489

Try to solve it in two steps: Find the speed of the ball as it leaves the gun. Then find the height that it reaches after it becomes a projectile. 


#3
Apr2410, 06:38 PM

P: 46

Well then I guess I have no idea what i'm doing then...A little more help please
K_i + U_i + S_i = K_f + U_f + S_f K = .5mv^2 U = mgh S = .5kx^2 What is the muzzle velocity? How to find the height that it reaches? 


#4
Apr2510, 09:36 AM

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P: 41,489

Ball Launched from spring gun (max height)



#5
Apr2510, 02:42 PM

P: 46

Could someone please solve the problem for me.....BTW this is NOT a homework problem....It is a final test review problem....I learn differently than most.....By looking at the steps involved in solving the problem along with the answer I can use deductive reasoning as to why those methods were used....TIA



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