5. Diana, a 65 kg volleyball player, lands on her feet after scoring an ace serv

Question

5. 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?

Explanation / Answer

Impulse is given by the formula J = F t = change in momentum

J = mv-mu

where m is mass

u is intial velocity

v ia final veloicity

so

part A : Impulse J = 65*(3-(-4) = 455 Kgm/s

part B : Avg force = J/t = 455/0.65 = 700 N

PartC C : Net force on Diana - 700 N

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