Answer
The direction of the impulse on the ball is at an angle of $~~59.8^{\circ}~~$ above the positive x axis.
Work Step by Step
In part (a), we found that $\Delta p_y = 5.065~kg~m/s$ and $\Delta p_x = 2.949~kg~m/s$
We can find the direction of the impulse on the ball relative to the positive x axis:
$tan~\theta = \frac{5.065~kg~m/s}{2.949~kg~m/s}$
$\theta = tan^{-1}~(\frac{5.065~kg~m/s}{2.949~kg~m/s})$
$\theta = tan^{-1}~(1.7175)$
$\theta = 59.8^{\circ}$
The direction of the impulse on the ball is at an angle of $~~59.8^{\circ}~~$ above the positive x axis.
