The travel of an acrobat through space can be modeled as the travel of a particle at the acrobat's center of mass, as we will study in a later chapter. The components of the displacement of an acrobat's center ofmass from the beginning to the end of a certain trajectory are described by the equationsX = 0 + (11.9 m/s) (cos(18.5°))Tf0.130 m = 0.650 m + (11.9 m/s)(sin(18.5°))T(9.80 m/s²)T/²where T, is in seconds and is the time it takes the acrobat to travel from the takeoff site to the landing point.(a) Identify the acrobat's position (in vector notation) at the takeoff point (in m). (Let the x- and y-direction be along the horizontal and vertical direction, respectively.)=(b) Identify the vector velocity at the takeoff point. (Enter the magnitude in m/s and the direction in degrees counterclockwise from the +x-axis.)m/smagnitudedirection° counterclockwise from the +x-axis(c) How far (in m) did the acrobat land from the takeoff point?Need Help?mmRead It

Given that:xf=0+11.9 m/s (cos18.5°)Tf ....(1)0.130m=0.650m+11.9 m/s (sin18.5°)Tf-12(9.80 m/s2)Tf2…

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