Chapter 4, Problem 4.26P

The motion of a human body through space can be modeled as the motion of a particle at the body’s cen­ter of mass as we will study in Chapter 9. The compo­nents of the displacement of an athlete’s center of mass from the beginning to the end of a certain jump are described by the equations

x_j = 0 + (11.2 m/s) (cos 18.5°)t 0.360 m = 0.840 m + (11.2 m/s)(sin 18.5°)t - $\frac{1}{2}$(9.80 m/s²)t² where t is in seconds and is the time at which the ath­lete ends the jump. Identify (a) the athlete’s position and (b) his vector velocity at the takeoff point. (c) How far did he jump?