Math Problem Statement

Austin (61.9 kg body mass) is recovering from knee surgery and is doing testing to be cleared to return to sport. One of the evaluations is a vertical jump on a force plate which gives ground reaction force. The data show that they stood motionless for 1 s then initiated the counter-movement jump. Vertical ground reaction force was reduced to 76.9% of body weight for 0.39 s during the counter movement. Vertical ground reaction force then increased to 186% of body weight for 0.25 s. What was the net impulse (impulse greater than what would be needed to simply support body weight). Give your answer to one decimal place.

Solution

To calculate the net impulse exerted during the jump, we can break the problem into two phases: the downward counter-movement phase (where the force is reduced below body weight) and the upward jump phase (where the force exceeds body weight). The net impulse is the sum of the impulses during these phases relative to the force needed to support Austin's body weight.

Step 1: Calculate Body Weight

Austin's mass mm is 61.9 kg, and gravitational acceleration gg is approximately 9.81m/s29.81 \, \text{m/s}^2.

Body weight=mg=61.9×9.81=607.239N\text{Body weight} = m \cdot g = 61.9 \times 9.81 = 607.239 \, \text{N}

Step 2: Impulse During Counter-Movement Phase

During the counter-movement phase, the force was reduced to 76.9% of Austin’s body weight for 0.39 s.

  • Force during counter-movement = 0.769×607.239N=467.179N0.769 \times 607.239 \, \text{N} = 467.179 \, \text{N}
  • Force needed to support body weight = 607.239 N

The impulse during this phase can be calculated as:

Impulsedown=(Force appliedBody weight)×Time\text{Impulse}_{\text{down}} = (\text{Force applied} - \text{Body weight}) \times \text{Time} Impulsedown=(467.179607.239)×0.39=140.06×0.39=54.62Ns\text{Impulse}_{\text{down}} = (467.179 - 607.239) \times 0.39 = -140.06 \times 0.39 = -54.62 \, \text{Ns}

Step 3: Impulse During Upward Phase

During the upward phase, the vertical ground reaction force increased to 186% of body weight for 0.25 s.

  • Force during upward phase = 1.86×607.239N=1129.46N1.86 \times 607.239 \, \text{N} = 1129.46 \, \text{N}
  • Force needed to support body weight = 607.239 N

The impulse during the upward phase is:

Impulseup=(1129.46607.239)×0.25=522.221×0.25=130.56Ns\text{Impulse}_{\text{up}} = (1129.46 - 607.239) \times 0.25 = 522.221 \times 0.25 = 130.56 \, \text{Ns}

Step 4: Calculate Net Impulse

The net impulse is the sum of the impulses from both phases:

Net impulse=Impulsedown+Impulseup=54.62+130.56=75.94Ns\text{Net impulse} = \text{Impulse}_{\text{down}} + \text{Impulse}_{\text{up}} = -54.62 + 130.56 = 75.94 \, \text{Ns}

Final Answer:

The net impulse is approximately 75.9 Ns.


Do you want further details or have any questions?
Here are 5 related questions for further exploration:

  1. How does impulse relate to momentum in this scenario?
  2. What effect would increasing the mass have on the net impulse?
  3. How could the duration of the counter-movement phase affect the net impulse?
  4. What role does the force plate play in biomechanical analysis?
  5. How can Austin improve his vertical jump performance based on this data?

Tip: Net impulse can also be viewed as the excess or deficit of force applied compared to just supporting body weight, making it crucial for understanding motion changes.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Impulse
Force
Newton's Laws of Motion
Biomechanics

Formulas

Body weight = m * g
Impulse = (Force applied - Body weight) * Time
Net Impulse = Impulse_down + Impulse_up

Theorems

Newton's Second Law of Motion (F = ma)

Suitable Grade Level

Grades 11-12 (or university-level biomechanics)