The path of a gymnast through space can be modeled as the path of a particle...

The path of a gymnast through space can be modeled as the path of a particle at the gymnast's center of mass, as we will study in a later chapter. The components of the displacement of a gymnast's center of mass from the beginning to the end of a certain trajectory are described by the equations xf = 0 + (10.3 m/s)(cos(18.5°))Tf 0.380 m = 0.640 m + (10.3 m/s)(sin(18.5°))Tf − 1 2 (9.80 m/s2)Tf2 where Tf is in seconds and is the time it takes the gymnast to travel from the takeoff site to the landing point. (a) Identify the gymnast'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.) r = m (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.) magnitude Incorrect: Your answer is incorrect. Compare generic expressions for either the x or y-components of the position of an object undergoing two-dimensional motion with the analogous expressions given in the statement of the problem. m/s direction ° counterclockwise from the +x-axis (c) How far (in m) did the gymnast land from the takeoff point?

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genius_generous answered 1 year ago

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