Ball Launched from spring gun (max height)

AI Thread Summary
A ball is launched from a spring gun, and the goal is to determine its maximum height above the ground. The discussion emphasizes using conservation of energy to find the ball's muzzle velocity and subsequent height after it leaves the gun. Key equations include kinetic energy (K = 0.5mv^2), gravitational potential energy (U = mgh), and spring potential energy (S = 0.5kx^2). The initial attempts to solve the problem incorrectly equate velocity and force, highlighting the need for a proper understanding of the concepts involved. The final takeaway is that solving the problem involves calculating the speed at launch and applying projectile motion principles thereafter.
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Homework Statement



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.

Homework Equations



(k(x^2)/2)=mgh

?? x = mgcos(theta)

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.
 
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jti3066 said:
(k(x^2)/2)=mgh
Here x is a velocity. You can use conservation of energy to find the additional height reached after the ball leaves the gun, if x is the vertical component of the speed at that point. Note that h will not be the height above the ground.

?? x = mgcos(theta)
You can't set a velocity (x) equal to a force.

The Attempt at a Solution



Plugging in "x" I get: h=(k(mgcos(theta))^2)/2mg

Is this correct?
No. Note that you haven't even used the spring constant or the amount the spring was compressed.

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.
 
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?
 
jti3066 said:
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?
To find the muzzle velocity, use conservation of energy as you outlined above. Hint: What's the change in height of the ball as it travels through the gun?

How to find the height that it reaches?
Once the ball leaves the gun it's a projectile. Solve that part like any other projectile motion problem.
 
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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