Question

A diver makes 1.0 revolutions on the way from a 9.2-m-high platform to the water. Assuming zero initial vertical velocity, find the diver's average angular velocity during a dive.

Answer 1

Answer:

Average angular velocity ≈ 4.59 rad/s

Explanation:

Using the equation of motion,

H = ut + (1/2)t² ............................ equation 1.

Where H= height, u = initial velocity(m/s), g = acceleration due to gravity(m/s²), t = time(s)  u= 0 ∴ ut =0

H =(1/2)gt².................................... equation 2.

making t² the subject of the relation in equation 2,

∴ t² = 2H/g

Where H = 9.2 m, g= 9.8 m/s

∴ t² = ( 2×9.2)/9.8

t = √(2 × 9.2/9.8) = √(18.4/9.8)

 t = 1.37 s.

The average angular velocity = θ/t

Where θ = is the number of revolution that the diver makes, t  = time

           θ = 1 rev.

Since 1 rev = 2π (rad)

           t = 1.37 s

 Average angular velocity = 2π/t

π = 3.143

 Average angular velocity = (2×3.143)/1.37 = 6.286/1.37

   Average angular velocity ≈ 4.59 rad/s