Answer
The spring launches the ball at a speed of $~~12~m/s$
Work Step by Step
We can find the vertical component of the velocity as the ball leaves the muzzle:
$v_{yf}^2 = v_{y0}^2+2a_y~h$
$0 = v_{y0}^2+2a_y~h$
$v_{y0}^2 = -2a_y~h$
$v_{y0} = \sqrt{-2a_y~h}$
$v_{y0} = \sqrt{-(2)(-9.8~m/s^2)(1.83~m)}$
$v_{y0} = 6.0~m/s$
We can find the speed at which the spring launches the ball:
$v_{0y} = v_0~sin~\theta$
$v_0 = \frac{v_{0y}}{sin~\theta}$
$v_0 = \frac{6.0~m/s}{sin~30^{\circ}}$
$v_0 = 12~m/s$
The spring launches the ball at a speed of $~~12~m/s$
