Calculating Muzzle Velocity of a Spring-Loaded Toy Gun

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Homework Statement



A spring-loaded toy gun is used to shoot a ball of mass m=1.5 kg straight up in the air. The spring has spring constant k=6.7N/m. If the spring is compressed a distance of 25.0 centimeters from its equilibrium position y=0 and then released, the ball reaches a maximum height (measured from the equilibrium position of the spring). There is no air resistance, and the ball never touches the inside of the gun. Assume that all movement occurs in a straight line up and down along the y axis.

Find the muzzle velocity of the ball (i.e., the velocity of the ball at the spring's equilibrium position ).


Homework Equations



1/2mv^2

mgy

U_i=K_f

The Attempt at a Solution



I came up that the final Kinetic energy was 208805 and that equaled 1/2mv^2 and attempted to solve for v and it was wrong. I'm having an error when trying to caulculate the final kinetic energy.
 
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[tex]F_s = kx[/tex]
[tex]E_s = 1/2 k x^2[/tex]

You'll need to use at least one of those two formulas as well, and realize that mechanical energy (sum of potential and kinetic) is conserved throughout.
 
Equations have = signs!

What does U_i=K_f mean?

What is the energy stored in a compressed spring (a.k.a SPE)?
 
Sorry for not being clear. "U_i=K_f" does have an = sign. "1/2mv^2" and "mgy" do not.

What does U_i=K_f mean?

What is the energy stored in a compressed spring (a.k.a SPE)?
 
Also, why do you want to use [tex]U_g = mgh[/tex]? At what position of the ball are you interested in its speed?
 
Last edited:
the spring energy is converted to kinetic energy.. and we are assuming there is no loss of energy due to minimal air resistance; therefore, set the kinetic energy= spring energy.
Note-- change 25cm to meters.