Diana, a 65 kg volleyball player, lands on her feet after scoring an ace serve t

Question

Diana, a 65 kg volleyball player, lands on her feet after scoring an ace serve then immediately jumps up again to celebrate her serve. The instant before her feet first touch the floor after the serve, her velocity is 3 m/s downward. Her velocity is 4 m/s upward when her feet leave the floor 0.65 s later. (a) What is the impulse exerted on Diana during the 0.65 s she is in contact with the ground?; (b) What is the average net force exerted on Diana during this 0.65 s?; and (c) What is the average reaction force exerted upward by the floor on Diana during the 0.65 s? (4 pts)

Please show all work and use correct signfiicant figures

Explanation / Answer

initial downward momentum =mu=65*3 = -195j

final upward momentum =mv =65*4 =260j

a) impulse = change in momentum =260+165 =455 m/s

b)

average net force exerted on Diana during this 0.65 s =impulse /time =455/0.65 =700 N

c)

average reaction force exerted upward by the floor on Diana is equal to average net force exerted on Diana

so

average reaction force exerted upward by the floor on Diana during the 0.65 s =700N

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