KE and U Energy in Spring Gun Question

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SUMMARY

The discussion focuses on the mechanics of a child's spring gun, specifically analyzing the kinetic energy (KE) and gravitational potential energy (U) involved in launching a 50 g ball. The spring constant is given as 730 N/m, and the ball reaches a maximum height of 1.75 m when launched at an angle of 33°. The user seeks clarification on the application of energy equations, particularly regarding the final velocity (Vf) at the peak height and the calculation of the spring's initial compression distance using the formula 1/2 k x^2 = 1/2 mv^2.

PREREQUISITES
  • Understanding of kinetic energy (KE) and gravitational potential energy (U)
  • Familiarity with Hooke's Law and spring constant (k)
  • Basic knowledge of projectile motion and angles
  • Ability to manipulate and solve algebraic equations
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  • Calculate the initial velocity of the ball using energy conservation principles
  • Explore the relationship between angle of launch and maximum height in projectile motion
  • Investigate the effects of spring constant on compression distance and launch speed
  • Learn about energy transformations in mechanical systems
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Students studying physics, educators teaching mechanics, and hobbyists interested in the physics of spring-loaded devices.

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The spring in the muzzle of a child's spring gun has a spring constant of 730 N/m. To shoot a ball from the gun, first the spring is compressed and then the ball is placed on it. The gun's trigger then releases the spring, which pushes the ball through the muzzle. The ball leaves the spring just as it leaves the outer end of the muzzle. When the gun is inclined upward by 33° to the horizontal, a 50 g ball is shot to a maximum height of 1.75 m above the gun's muzzle. Assume air drag on the ball is negligible. (a) At what speed does the spring launch the ball? (b) Assuming that friction on the ball within the gun can be neglected, find the spring's initial compression distance.


i know how to solve it but my question is:

1/2 m (Vf^2 - Vi^2)= - mgh

the teacher put vf^2 = Vi cos 33

but I am wondering isn't Vf= 0 when it reaches the top??

Does KE and U energy have a components like forces??
 
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and for part b)
1/2 k x^2 = 1/2 mv^2
correct??
 
anyone??
 

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