Calculating Spring Compression and Muzzle Velocity in Projectile Motion

[bacon]

Homework Statement

The spring of a gun has a spring constant, k, of 4.0lb/in. When the gun is inclined upward by 30 degrees to the horizontal, a 2.0 oz ball is shot to a height of 6.0 ft above the muzzle of the gun. (a) What was the muzzle speed of the ball? (b) By how much must the spring have been compressed initially?
No questions on part (a). I can get the answer of 39.2 f/s.
The answer to part (b) is 3.4 in. I am consistently getting an answer of 4.2 in.

Homework Equations

k=4.0 lb/in = 48 lb/f
m= "2.0 oz" = 2.0 x $$(1.94)10^{-3}$$ slugs/oz= 3.88 x $$10^{-3}$$slugs
$$E_{k}=\frac{1}{2}mv^{2}$$

Potential energy contained in the spring:
$$U=\frac{1}{2}kx^{2}$$

The Attempt at a Solution

The energy of the ball as it leaves the muzzle comes from the potential energy released as the spring decompresses. Therefore,

$$U=E_{k}$$
Solving for x gives:

$$x=v\sqrt{\frac{m}{k}}$$

and plugging in values for the variables:

$$x=39.2\sqrt{\frac{(3.88)10^{-3}}{48}}=.35f=.35(12)=4.2in$$

which is apparently the wrong answer.
Any help would be appreciated.
Thank you.

 

[diazona]

Your method looks right, and I get the same answer you do, 4.2 inches. Either I'm missing something, or the given answer is wrong. (I double-checked your answer for part (a) too, just in case, but I also get 39.2 ft./s)

 

[bacon]

Thanks, Diazona,
Not only do I get 39.2f/s in part (a), it is the answer in the back of the book. So I feel fairly sure that part is correct. I am also starting to think the answer in the book for (b) is wrong but wanted to make sure I wasn't missing something.