Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled
Let
α = the angular acceleration
ω = the final angular velocity
The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²
The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s
If the thrower's arm is r meters long, the tangential velocity of release will be
v = 13.963r m/s
Answer: 13.963 rad/s
Answer:

Explanation:
given,
weight of swimmer = 510 N
length of ledge, L = 1.75 m
vertical height of the cliff, h = 9 m
speed of the swimmer = ?
horizontal velocity of the swimmer should be that much it can cross the wedge.
distance = speed x time
d = v_x × t
1.75 = v_x × t ........(1)
now,time taken by the swimmer to cover 9 m
initial vertical velocity of the swimmer is zero.
using equation of motion for time calculation


t² = 1.938
t = 1.39 s
same time will be taken to cover horizontal distance.
now, from equation 1
1.75 = v_x × 1.39

horizontal speed of the swimmer is equal to 1.26 m/s