Can someone give me a clue for this problem?

Question

The launching mechanism of a toy gun consists of a spring of unknown spring constant, as shown in Figure P5.33a. If its spring is compressed a distance of 0.160 m and the gun fired vertically as shown, the gun can launch a 23.0 g projectile to a maximum height of 24.0 m above the starting point of the projectile.

(a) Neglecting all resistive forces, determine the spring constant.
N/m
(b) Neglecting all resistive forces, determine the speed of the projectile as it moves through the equilibrium position of the spring (where x = 0), as shown in Figure 5.33b.
m/s

Answer 1

Here the potential energy of the spring is converted to the potential energy of the projectile.

Ps=Pe
(1/2)kx=mgh
a) k= 2mgh/x
k=2 x 0.023 x 9.81 x 24.0/ 0.160=
k=67.7 N/m

b) In this case as soon as the Ps is exhausted it is compleatly trasformeed into kinetic energy Ke however at the very to Ke becomes Pe.
So Ke=Pe

(1/2) mV^2=mgh
V=sqrt(2gh)
V=sqrt(2 x 9.81 x 24.0)=
V=21.7 m/s

Answer 2

Actually, Edward forgot to square the compressed distance (x) in his answer for a. The following are the steps to solve:
a)
Ps=Pe
(kx^2)/2=mgh
k=2mgh/x^2
k=(2*.023*24)/(.16^2)
k=405.4289063N/m

b)
The Ps will equal Ke as the Potential energy of the spring will have been completely exhausted and transformed entirely to kinetic energy. However, one can't compute with this as x=0 will make Ps 0, and kinetic energy isn't 0 at the equilibrium. So, we must compute with the potential energy as the law of conservation of energy states no energy can be lost, simply transformed. So, we compute as follows:

mgh=(mv^2)/2
2mgh=mv^2
2gh=v^2
v=sqrt(2gh)
v=sqrt(2*9.81*24)
v=21.69976958m/s

Answer 3

find the velocity(v) to reach that max. height .

equate energy of the spring in the compressed position to 1/2 m(v^2) , where m is the mass of the projectile.