Jones set a new record in the standing long jump. During his record setting perf

Question

Jones set a new record in the standing long jump. During his record setting performance he took off with a velocity of 6.73 m/s at an angle of 27o relative to the right horizontal. Assume the jumper’s center of mass height at take-off and landing were the same.

a. What were his horizontal and vertical velocities at take-off?

b. What was his total time in the air?

c. How high above the take-off position did Byron’s center of mass rise?

d. What was the horizontal displacement of the Byron center of mass?

Explanation / Answer

initial velocity U = 6.73 m/s

angle = 27 deg

a) horizontal velocity Ux= 6.73*cos27 = 5.996 m/s

vertical velocity Uy= 6.73* sin27 = 3.055 m/s

b) total time = 2Uy/g = 2*5.996/9.81 = 1.22 s

c) height = Uy^2/2g = 5.996*5.996/2*9.81 = 1.832 m

d) horizontal displacement = Ux*t = 3.055*1.22 = 3.72

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