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Peter, a 100 kg basketball player, lands on his feet after completing a slam dunk and then immediately jump up again to celebrate his basket. When his feet first touch the floor after the dunk,
his velocity is 5 m/s downward; when his feet leave the floor 0.50 s later, as he jumps back up,
his velocity is 4 m/s upward. What is the average reaction force exerted upward by the floor on Peter during this 0.50 s?

Respuesta :

kweenisbah

Answer:

1800 N

Solution:

Impulse = mΔv = m * (u - v) .

 here m = 100 kg

          u = 4 m/s

          v = -5 m/s

  impulse = 100 x ( 4 - ( -5 ) ) = 900 Kg m/s .

Average reaction Force ( Favg ) = impulse / Δt

Average reaction Force ( Favg )  = 900kg·m/s / 0.5s

Average reaction Force ( Favg )  = 1800 N

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