Chapter 12 - Kinematics of a Particle - Section 12.6 - Motion of a Projectile - Problems - Page 50: 82

$v=\sqrt{c^2k^2+b^2}$ $a=ck^2$

$x=c\sin kt$ $\dot{x}=ck\cos kt$ $\ddot{x}=-ck^2\sin kt$ $y=c\cos kt$ $\dot{y}=-ck\sin kt$ $\ddot{y}=-ck^2\cos kt$ $z=h-bt$ $\dot{z}=-b$ $\ddot{z}=0$ $v=\sqrt{(ck\cos kt)^2+(-ck \sin kt)^2+(-b)^2}=\sqrt{c^2k^2+b^2}$ $a=\sqrt{(-ck^2\sin kt)^2+(-ck^2\cos kt)^2+0}=ck^2$