Spring constant physics question?

Question

I'm stuck on this problem. I don't know how to do it! Please show the steps.
Thank you!

The launching mechanism of a toy gun consists of a spring of unknown spring constant. If its spring is compressed a distance of 0.120 m and the gun fired vertically as shown, the gun can launch a 20.0 g projectile to a maximum height of 20.0 m above the starting point of the projectile.

(a) Neglecting all resistive forces, determine the spring constant.

(b) Neglecting all resistive forces, determine the speed of the projectile as it moves through the equilibrium position of the spring (where x = 0)

Answer 1

a) E.P.E=G.P.E
E.P.E=1/2XforceXextension
Force = Kx
E.P.E= 1/2X(k)X(Extension)^2

G.P.E=MXgrav acc. X height
G.P.E=(20/1000)(9.8)(20.0)=3.92 J
3.92 = 1/2(K)(0.120)^2
K(spring constant ) =544.4444444 Nm^-1

b) V^2=2as+U^2
s=20.0
a=9.8
u = 0 (because at max height velocity is 0)
V^2=2(9.8)(2)
V^2=392
V=19.79898987 ms^-1

Answer 2

Lets go about the systematically. 600g=.6kg (mass of object) F=Ma (force of gravity equals mass times gravitational acceleration) F=(.6)(9.8)=5.88N k=24N/M (spring constant) F=kx (Force of spring, must be equal to force of gravity on object) 5.88N=(24)(x) x=.25m

Answer 3

(1/2)kx^2 = mgh
a)
k = 2mgh/x^2
k ≈ (2)(0.02)(9.80665)(20.0) / 0.12^2
k ≈ 544.8139 N/m ≈ 545 N/m
b)
(1/2)mv^2 = mgh
v^2 = 2gh
v ≈ √((2)(9.80665)(20.0)
v ≈ 19.80571 m/s ≈ 19.8 m/s