### Kinematics

The velocity at the beginning has two components:
$v_{0x} = (15\frac{m}{s})cos(20^{o})=14\frac{m}{s}$
$v_{0y} = -(15\frac{m}{s})sin(20^{o})=-5.1\frac{m}{s}.$

(a)The horizontal displacement (distance) is $x=v_{x}t=(14\frac{m}{s})(2.3x)=32m.$

(b)The vertical displacement is
$y = \frac{-gt^{2}}{2}+v_{0y} = -(9.8\frac{m}{s^{2}})\frac{(2.3s)^{2}}{2}+(-5.1\frac{m}{s})(2.3s)=-38m.$