Answer
The direction of the average force 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 vertical component of the average force:
$F_y = \frac{5.065~kg~m/s}{2.00\times 10^{-3}~s}$
$F_y = 2530~N$
We can find the horizontal component of the average force:
$F_x = \frac{2.949~kg~m/s}{2.00\times 10^{-3}~s}$
$F_x = 1470~N$
We can find the direction of the average force on the ball relative to the positive x axis:
$tan~\theta = \frac{2530~N}{1470~N}$
$\theta = tan^{-1}~\frac{2530~N}{1470~N}$
$\theta = tan^{-1}~(1.72)$
$\theta = 59.8^{\circ}$
The direction of the average force on the ball is at an angle of $~~59.8^{\circ}~~$ above the positive x axis.
