Tamia L.
asked • 02/02/21
When a 0.622 kg basketball hits the floor, its velocity changes from 4.23 m/s down to 3.85 m/s up. What impulse was given to the ball?
Dr Gulshan S.
answered • 02/03/21
Physics Teaching is my EXPERTISE with assured improvement
When a 0.622 kg basketball hits the floor, its velocity changes from 4.23 m/s down to 3.85 m/s up. What impulse was given to the ball?
Impulse = Change in momentum
= Mass * Change in Velocity = 0.622 (4.23 - 3.85) = 0.23636 Kgm/s
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