When a 0.622 kg basketball hits the floor, its velocity changes from \( 4.23 \mathrm{~m} / \mathrm{s} \) down to \( 3.85 \mathrm{~m} / \mathrm{s} \) up. If the average force was 72.9 N , how much time was it in contact with the floor? (Unit = s) Remember: up is + , down is - Sthint
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- Mass of the basketball, \( m = 0.622 \) kg - Initial velocity, \( v_i = -4.23 \) m/s (downward, hence negative) - Final velocity, \( v_f = 3.85 \) m/s (upward, hence positive) - Average force, \( F = 72.9 \) N Show more…
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A basketball with a mass of 0.62 kg falls vertically to the floor where it hits with speed of -6 m/s. (We take the positive direction to be upward here). The ball rebounds, leaving the floor with a speed of 4.5 m/s. If the ball is in contact with the floor for 0.04 s, what is the average force of the ball on the floor?
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