Asked by Eden

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=667 N/m. If the spring is compressed a distance of 25.0 centimeters from its equilibrium position 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 y=0).

Answers

Answered by bobpursley
Energy in spring= 1/2 k .25^2
gravitational energy gained by ball going to equilibrium position 1.5*g*.25

KE at equilibrium= 1/2k .25^2-1.5*g*.25
and of course, KE is 1/2 1.5*v^2
solve for v.
Answered by Shandon
Ui=mgh+.5kh^2
ui=(1.5)(9.81)(-.25)+(.5(667)(-.25)^2)

ki=0 Ui=17.165 Kf=.5mv^2

ki+Ui=.5mv^2 solve for v

v=4.78 m/s

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