Answer
The launch angle is $~~48.2^{\circ}$
Work Step by Step
The minimum value of momentum is $4.0~kg~m/s$. This must occur when the ball reaches maximum height. Therefore the horizontal component of momentum must be $p_x = 4.0~kg~m/s$
The initial momentum was $6.0~kg~m/s$
We can find the launch angle:
$p_x = p_0~cos~\theta$
$cos~\theta = \frac{p_x}{p_0}$
$\theta = cos^{-1}~(\frac{p_x}{p_0})$
$\theta = cos^{-1}~(\frac{4.0~kg~m/s}{6.0~kg~m/s})$
$\theta = 48.2^{\circ}$
The launch angle is $~~48.2^{\circ}$.
