Ball Launched from spring gun (max height)

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Homework Help Overview

The problem involves a ball of mass m launched from a spring gun at an angle theta, with the goal of determining the maximum height h the ball reaches above the ground. The spring has a spring constant k and is compressed to a length d before launch.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using conservation of energy to relate the spring's potential energy to the ball's kinetic and gravitational potential energy. There are questions about the correct interpretation of variables and the steps needed to find the muzzle velocity and subsequent height of the projectile.

Discussion Status

Some participants are exploring the relationship between the spring's compression and the energy transformations involved. Guidance has been offered on how to approach finding the muzzle velocity and treating the ball as a projectile after it leaves the gun. However, there is no explicit consensus on the correct method or solution yet.

Contextual Notes

One participant mentions that this is a final test review problem, indicating that it may not follow typical homework constraints. There are indications of varying levels of understanding among participants regarding the application of energy conservation principles.

jti3066
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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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