angular displacement

Question

An athlete is performing a vertical jump. The first jump indicates max knee flexion prior to the jump. The knee flexion angle is 86 degrees and the time is 2.25 seconds. At peak height, the knee flexion angle is 178 degrees at 2.85 seconds.

a) what is his angular displacement from maximum knee flexion prior to the jump to the peak height of his jump?

b) what is his angular velocity from max knee flexion to the peak height of his jump?

Answer 1

178° - 86° = 92° (this is the angular displacement in degrees)

92° x (π radians)/(180°) = 1.6057 radians = 1.61 radians (this is the angular displacement in radians)

Δt = 2.85 - 2.25 = 0.60 seconds

angular velocity (ω) is Δθ/Δt = (1.6057 radians)/(0.60 seconds) = 2.68 radians/sec