I have no clue how to do this, please help. The motion of a human body through s

Question

I have no clue how to do this, please help.

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 + (10.6 m/s)(cos 18.5 degree)t 0.260 m = 0.620 m + (10.6 m/s)(sin 18.5 degree)t - 1/2(9.80 m/s2)t2 where t is in seconds and is the time at which the athlete ends the jump. Identify the athlete's position at the takeoff point. (Let the x and y-direction be along the horizontal and vertical direction respectively.) Identify his vector velocity at the takeoff point, magnitude direction m/s above the x axis How far did he jump?

Explanation / Answer

(a)

The displacement is,

xf = 0 + (10.6 m/s)(cos 18.5°)t
   0.260 m = 0.620 m + (10.6 m/s)(sin 18.5°)t - ½(9.80 m/s2)t2

The athlete's position at the takeoff point is,
t = 0, x = 0, y = 0.620 m

So, the athlete's position at the takeoff point is,

(0 i + 0.620 j) m

(b)

The athlete’s vector velocity at the takeoff point is,

t = 0,

     dx/dt = 10.6cos(18.5) m/s

= 10.05 m/s

     dy/dt = 10.6sin(18.5) m/s

              = 3.36 m/s

Vector velocity is,

v= (10.6 i + 3.36 j) m/s
magnitude = 11.1 m/s
direction 17.6° above the x axis

(c)

Solve the equation.

0.260 m = 0.620 m + (10.6 m/s)(sin 18.5°)t - ½(9.80 m/s2)t2
   

and get t = 0.0.94 s

So,

x = 10.6 cos(18.5)* 0.094 s

= 0.944 m

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