Answer
(a) $t=0.25\text{ s}$;
(b) $h=2.1\text{ m}$.
Work Step by Step
(a) The horizontal component of the velocity of the ball is constant and equal to $v_0\cos32^\circ$. The ball has to close $3.8\text{ m}$ of horizontal distance so the time is
$$t=\frac{3.8\text{ m}}{18\text{ m/s }\cdot \cos32^\circ}=0.25\text{ s}.$$
(b) The ball moves vertically with the initial vertical component of the velocity that is equal to $v_0\sin32^\circ$ and points upwards and has a downward acceleration equal to $g$. The height of the ball when it hits the wall is equal to its $y$ coordinate at previously calculated time $t$ so we have
$$h=v_0t\sin32^\circ-\frac{1}{2}gt^2=18\text{ m/s }\cdot0.25\text{ s }\cdot\sin32^\circ-\frac{1}{2}9.81\text{ m/s}^2\cdot(0.25\text{ s})^2=2.1\text{ m}.$$
