How far did he jump? The motion of a human body through space can be modeled as

Question

How far did he jump?

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 equation:s xf0+(11.7 m/s)(cos 18.5o)t 0.320 m = 0.690 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. (a) Identify the athlete's position at the takeoff point. (Let the x and y-direction be along the horizontal and vertical direction respectively.) (b) Identify his vector velocity at the takeoff point. magnitude 11.7 direction 18.5 above thex axis (c) How far did he jump? Need Help? ReadIt

Explanation / Answer

(c)

along y direction

0.32 = 0.69 + (11.7*sin18.5*t) - (1/2)*9.8*t^2

time of jump t = 0.85 s

along x direction

xf = 0 + 11.7*cs18.5*t)

xf = 11.7*cos18.5*0.85

xf = 9.43 m <<<<<=========ANSWER

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