Answer
The spring's initial compression distance is $~~11~cm$
Work Step by Step
In part (a), we found that the spring launches the ball at a speed of $~~12~m/s$
We can use conservation of energy to find the spring's initial compression distance:
$U_e = K$
$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$
$x^2 = \frac{mv^2}{k}$
$x = \sqrt{\frac{mv^2}{k}}$
$x = \sqrt{\frac{(0.057~kg)(12~m/s)^2}{700~N/m}}$
$x = 0.11~m$
$x = 11~cm$
The spring's initial compression distance is $~~11~cm$
