### Kinematics

The path of the ball is a vector and we break it into its different
components.

(a) We break up the initial velocity of the ball:
$v_{0x} = (25.3\frac{m}{s})cos(42.0^{o})=18.8\frac{m}{s}\\ v_{0y} = (25.3\frac{m}{s})sin(42.0^{o})=16.9\frac{m}{s}$
The ball is in the air for a time $t=x/v_{x}=(21.8m)/(18.8m/s)=1.16s$
before it hits the wall.

(b)$y=-gt^{2}/2+v_{oy}t=-(4.91m/s^{2})(1.16s)^{2}+(16.9m/s)(1.16s)=13.0m.$

(c)$v_{x}=v_{0x}=18.8m/s,\: v_{y}=-gt+v_{oy}=-(9.80m/s^{2})(1.16s)+(16.9m/s)=5.53m/s.$

(d) Since $v_{y}>0$, the ball is still going up and has not passed
the highest point yet.