What is the Spring Constant of a Gun Fired Ball?

[boogaaaaa]

Homework Statement

A spring loaded gun fires a 0.56g ball at an angle of 45 degrees from a horizontal lab bench at a height of 1.30m. The ball strikes the floor 5.64m horizontally (range) from the point of release. If the spring is compressed by 6.50 cm, what is the gun's spring constant

Homework Equations

1/2kx^2

1/2mv^2

The Attempt at a Solution

The final step is to sub the values for 1/2mv^2 = 1/2kx^2 and find the value for k.

Here's my diagram

http://img210.imageshack.us/img210/9712/diagramu.jpg

Cos 45 = Vx/V1
Vx = cos 45(v1)

Vx = Dx/t
(cos45)(v1) = dx/t
v1 = dx/(cos 45)t

From here I need to solve for time, but i have no idea how to

 

[nasu]

MG][/URL]

Cos 45 = Vx/V1
Vx = cos 45(v1)

Vx = Dx/t
(cos45)(v1) = dx/t
v1 = dx/(cos 45)t

From here I need to solve for time, but i have no idea how to

You cannot, you have two unknowns: v1 and t.
You need a second equation and this may be the equation of motion for the vertical (y) axis.

 

[boogaaaaa]

dy = (vy)(t) + (.5)(a)(t)
1.3 = v1(sin 45)t + .5(9.8)t^2

from here can I just sub in v1=dx/cos 45(t)?

If I do, I get the spring constant to be 9 (correct answer is 6) ;/

 

[nasu]

dy = (vy)(t) + (.5)(a)(t)
1.3 = v1(sin 45)t + .5(9.8)t^2

from here can I just sub in v1=dx/cos 45(t)?

If I do, I get the spring constant to be 9 (correct answer is 6) ;/

You are on the right track but your equation is not quite correct.
v1 is upward and the acceleration is downward so you need a minus sign there.
The complete equation would be
y=y0+volt-1/2 gt^2
with yo=1.3 and y=0 at t (final time), positive direction upward and origin at ground level.

 

[boogaaaaa]

Ok thank you, I got it now