Answer
(a) The direction of motion is $25^\circ$ above the horizontal. The final velocity magnitude is $v_f=16.8\text{ m/s}$.
(b) The ball has not reached the highest point.
Work Step by Step
We already calculated that the required time for the ball to reach the wall is $t=0.25\text{ s}$.
(a) Now we need to calculate the final values of the ball's velocity components:
$$v_{xf}=v_{x0}=v_0\cos32^\circ;\quad v_{yf}=v_{y0}-gt=v_0\sin32^\circ-gt.$$
Now, if $\theta$ is the angle between the ball's direction of motion and the horizontal, it holds:
$$\tan\theta=\frac{v_{yf}}{v_{xf}}=\frac{v_0\sin32^\circ-gt}{v_0\cos32^\circ}$$
which yields
$$\theta=\arctan\frac{v_0\sin32^\circ-gt}{v_0\cos32^\circ}=\arctan\frac{18\text{ m/s }\cdot\sin32^\circ-9.81\text{ m/s }\cdot0.25\text{ s}}{18\text{ m/s }\cdot\cos32^\circ}=25^\circ$$
and the direction is above the horizontal.
The magnitude of the velocity is found to be:
$$v_f=\sqrt{v_0^2-2gh}=\sqrt{18^2-2\cdot9.81\cdot2.1}\text{ m/s}=16.8\text{ m/s}.$$
(b) Since the direction of the ball's motion is still above the horizontal (the angle of the direction of motion with respect to the horizontal decreased from the initial $32^\circ$ down to $25^\circ$), the ball has not reached the highest point of its hypothetical parabolic trajectory. It will reach the highest point after it deflects off the wall.
