Biomechanics Problem- Impulse and projectile motion

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SUMMARY

The discussion focuses on solving a biomechanics problem related to impulse and projectile motion in a long jump scenario. The athlete's vertical ground reaction force and velocities at take-off are provided, with a weight of 850 N. To determine the jump height and distance, the impulse-momentum principle is essential, specifically using the equation Ft = mv. The optimal take-off angle for maximum horizontal distance is identified as 45 degrees.

PREREQUISITES
  • Understanding of basic linear kinematics equations
  • Familiarity with impulse-momentum theory
  • Knowledge of projectile motion principles
  • Ability to calculate weight and mass conversions (weight = mass × gravity)
NEXT STEPS
  • Study the impulse-momentum theorem in detail
  • Learn about projectile motion and optimal launch angles
  • Explore the effects of vertical and horizontal velocities on jump performance
  • Investigate methods to analyze ground reaction forces in biomechanics
USEFUL FOR

Students in biomechanics, sports scientists, coaches, and athletes looking to enhance their understanding of jump performance and the physics involved in athletic movements.

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Homework Statement


A long jump trial, the vertical ground reaction force of an athlete’s take-off
leg was measured as shown below. At foot strike, the horizontal velocity of the take -off leg was
8.00 m/s and the vertical velocity of the take-off leg was -2.30 m/s.

(Chart as Attachment)


a) Given that he weighed 850 N, how far and how high did he jump?
b) How could he improve his horizontal distance?
c) Discuss the kinetic factors that influence performance.


Homework Equations



Basic linear kinematics equations as well as Impulse:

Ft = mv



The Attempt at a Solution



I think I can use linear kinematics for the two distances, but my prof said that Impulse is key to solving it. Can anyone help with answer a in particular?
 

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The area under that graph is the impulse, but first I think you should subtract the force that simply opposes his weight. With the impulse and your equation, you can get his vertical velocity at take off (don't forget to get his mass by dividing their weight by gravity.)

He'll go furthest if he takes off at 45 degrees.

I have no idea how to interpret "horizontal velocity of take off leg". I reckon that ought to be zero otherwise he'd be slipping. Very badly worded question. You should flunk your prof.
 

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