The motion of a human body through space can be modeled as the motion of a parti

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 + (11.7 m/s)(cos 18.5°)t
0.200 m = 0.750 m + (11.7 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.

(b) Identify his vector velocity at the takeoff point.

direction

(c) How far did he jump?

Explanation / Answer

Here ,

b)

xf = 0 + (11.7 m/s)(cos 18.5°)t

xf = v * cos(theta) * t

comparing both equations

v = 11.7 m/s

theta = 18.5 degree

the velocity of 11.7 m/s at 18.5 degree

c)

0.200 m = 0.750 m + (11.7 m/s)(sin 18.5°)t 0.5*(9.80 m/s2)t^2

solving the equation for t

t = 0.885 s

distance = 11.7 * cos(18.5) * 0.885

distance = 9.82 m

the distance of jump is 9.82 m

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