A particle of mass M = 0.30 kg is dropped from a point that is at a height h = 2.10 m above the ground and horizontal distance s = 0.50 m from an observation point O, as shown in the figure. What is the magnitude of the angular momentum of the particle with respect to point O when the particle has fallen half the distance to the ground? (kg*m^2/s)
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We can use the equation of motion for this: v = sqrt(2*g*h/2) where g is the acceleration due to gravity (9.8 m/s^2) and h is the height (2.1 m). Plugging in the values, we get: v = sqrt(2*9.8*2.1/2) = sqrt(20.58) = 4.54 m/s Show more…
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