MUHAMMAD SHAKIR A. answered • 04/20/23
Online Physics tutor
a) The angular displacement from maximum knee flexion to the peak height of the jump is:
θ = θf - θi = 178° - 86° = 92°
Therefore, the athlete's angular displacement from maximum knee flexion prior to the jump to the peak height of his jump is 92 degrees.
b) To find the angular velocity from max knee flexion to the peak height of his jump, we need to know the initial and final angular velocities.
We can find the initial angular velocity using the first jump data. At maximum knee flexion, the athlete's leg is not moving, so the initial angular velocity is zero (ω₀ = 0).
To find the final angular velocity, we can use the formula:
ω² = ω₀² + 2αθ
where α is the angular acceleration, which we can assume to be constant.
The time interval for the jump is:
t = 2.85 s - 2.25 s = 0.60 s
The final knee flexion angle is 178 degrees. To convert this to radians, we multiply by (π/180):
θf = 178° x (π/180) = 3.11 radians
Similarly, the initial knee flexion angle is 86 degrees, or 1.50 radians:
θi = 86° x (π/180) = 1.50 radians
The angular displacement during the jump is:
θ = θf - θi = 3.11 - 1.50 = 1.61 radians
Substituting the known values into the equation for angular velocity:
ω² = ω₀² + 2αθ
we get:
ω² = 0 + 2(α)(1.61)
where α is the angular acceleration.
The angular acceleration can be found using the formula:
α = (ω - ω₀) / t
Since the initial angular velocity is zero:
α = ω / t
Substituting the known values:
α = (ω) / (0.60 s)
Solving for ω, we get:
ω = α x t = (2θ / t)^(1/2)
ω = (2 x 1.61 / 0.60)^(1/2) = 2.38 rad/s
Therefore, the athlete's angular velocity from max knee flexion to the peak height of his jump is 2.38 rad/s.
