Answer
The impulse given to the ball is $2.4~kg\cdot m/s$ in the horizontal direction away from the wall.
Work Step by Step
Note that only the horizontal component of velocity changes direction, so we only need to consider the horizontal change in momentum.
$F\cdot t = \Delta p = m \Delta v$
$F\cdot t = (0.060~kg)(2)\times (28~m/s)~cos(45^{\circ})$
$F\cdot t = 2.4~kg\cdot m/s$
The impulse given to the ball is $2.4~kg\cdot m/s$ in the horizontal direction away from the wall.
