Question

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 a later chapter. 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

xf = 0 + (10.5 m/s)(cos 18.5°)t

0.400 m = 0.610 m + (10.5 m/s)(sin 18.5°)t − ½(9.80 m/s2)t2

where t is in seconds and is the time at which the athlete ends the jump.
(a) Identify the athlete's position at the takeoff point. (Let the x and y-direction be along the horizontal and vertical direction resp

Answer 1

Find the value of t from the second equation
0.400 m = 0.610 m + (10.5 m/s)(sin 18.5°)t − ½(9.80 m/s2)t2
Solve the quadratic equation and we get t= 1.37 or -20.64 since time ‘t’ can’t be negative we take the positive value 1.37 plug the value of t in first equation to get
xf = 0 + (10.5 m/s)(cos 18.5°)t
xf= 13.52
not exactly sure if that answers the question