Question

Katelyn (55 kg) is practicing a drop jump in the biomechanics lab. She steps off a plyometrics box, lands on the force plate, and immediately jumps back up into the air. Right before she hits the force plate, her vertical velocity is 3.0 m/s downwards. After leaving the ground again, her vertical velocity is 3.5 m/s upwards. Katelyn was in contact with the ground for 0.4 seconds. (a) What was the impulse exerted on Katelyn when she was on the force plate

Answer 1

Answer:

J = 357.5 kg*m/s

Explanation:

• The impulse exerted on Katelyn when she was on the force plate, is equal to the change in her momentum, according to Newton's 2nd Law.
• Assuming as the positive direction the upward direction (coincident with the positive y-axis) we can express the initial momentum as follows:

       $p_{o} = m*v_{o} = 55 kg * (-3.0 m/s) (1)$

• By the same token, the final momentum is as follows:

       $p_{f} = m*v_{f} = 55 kg * (3.5 m/s) (2)$

• As we have already said, the impulse J is just equal to the change in momentum, i.e., the difference between (2) and (1):

      $J = p_{f} - p_{o} = m* (v_{f} -v_{o}) = 55 kg* (3.5m/s- (-3.0m/s)) = 357.5 kg*m/s (3)$