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A ball is projected at 20*sqrt(2) m/s at 45 deg above the horizontal. Find the angular velocity of the ball about the point of projection when it is at the highest point of its trajectory. (g = 10 m/s²)

1. 0.1 rad/s
2. 0.2 rad/s
3. 0.3 rad/s
4. 0.4 rad/s

Correct answer: 0.2 rad/s

Solution

At the highest point the speed is horizontal = u cos45 = 20 m/s. Using the geometry of the trajectory's top point, omega = v_perp / r works out to 0.2 rad/s.

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