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.

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