physics
posted by Liz on .
A spring gun (k = 27.1 N/m) is used to shoot a 51.1g ball horizontally. Initially the spring is compressed by 16.5 cm. The ball loses contact with the spring and leaves the gun when the spring is still compressed by 11.5 cm. What is the speed of the ball when it hits the ground, 1.41 m below the spring g

kx₁²/2 kx₂²/2 = mv(x)²/2
Find v²(x)
Vertical motion:
h=gt²/2 => t=sqrt(2h/g)
v(y)=gt.
v=sqrt{v²(x)+ v²(y)}