Dynamics/Kinematics - Jumper beats record jump

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SUMMARY

The jumper, weighing 60 kg, needs to exert a force greater than 1920 Newtons (N) to surpass the school record jump of 1.1 meters. The calculations involve using the equations of motion, specifically ma = ƩF and V² - V₀² = 2a(X - X₀). The jumper's initial velocity at the apex is determined to be 4.69 m/s, and the necessary acceleration during the jump is calculated to be approximately 21.996 m/s². The net force required for the jump is derived from the equation F(jump) = 60a + 600, leading to the final force requirement.

PREREQUISITES
  • Understanding of Newton's Second Law (ma = ƩF)
  • Familiarity with kinematic equations (V² - V₀² = 2a(X - X₀))
  • Basic knowledge of gravitational acceleration (g = -10 m/s²)
  • Ability to manipulate algebraic equations for solving physics problems
NEXT STEPS
  • Study the derivation and application of kinematic equations in projectile motion
  • Explore the concept of net force and its calculation in dynamic systems
  • Learn about the relationship between force, mass, and acceleration in real-world scenarios
  • Investigate the effects of different jumping techniques on performance outcomes
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Students in physics, athletic trainers, coaches, and anyone interested in biomechanics and performance optimization in sports.

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Homework Statement


The jumper is 60 kg and wants to beat the school's record jump, 1.1 m. She bends her knees a distance of .5 m before jumping. How many Newtons must Fjump be to for her to beat the record?

Homework Equations


ma = ƩF
V2 - V02 = 2a(X - X0)

My teacher stated I would just need these two equations, and that I would use the latter twice.

The Attempt at a Solution


Jumping up to the apex:
02 - 02 = (2)(-10)(1.1)
She jumps up from rest and the velocity of a mass at the apex is zero; right after she jumps, the only force acting on her is gravity, which is -10 m/s2. I assume 1.1 to be X because she can beat the previous runner's score if she jumps 1.1000001 m, essentially 1.1. So in my solution I could say "she has to jump with a force great than _X."

60a = Fjump - 600
I need acceleration from the previous step.

a) I don't know how to work in the .5 m into the equations. Would I add .5 to X? I would still get a nonsensical answer for step 1.
b) What is wrong with step 1's set up?
 
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The problem with step 1 is with your definition of "initial". The v0 in your equation is not the speed at the beginning of the jump. It's the launch speed: the speed at the end of the jump, once the jumper's legs are straight and she leaves the ground. And v is the final speed, the speed at the apex, which is 0. So, we have:

v0^2 = -2gh, where h = 1.1 m, the desired height of the jump.

Okay, that's all well and good, and from this we can find the launch speed needed.

How do we get the jump force? That's step 2, where we apply the kinematics equation again. It must be that the acceleration from the net force Fjump - weight, applied over a distance of 0.5 m, gives you a final v that is equal to v0 from step 1. So, in *step 2*, "initial" now means at the beginning of the jump. So v0 = 0, and v = the needed launch speed from step 1, and d = 0.5 m. From this you can solve for 'a', which gives you Fnet.
 
Cepheid variable!
---
Straight legs:
v0^2 = 2*-10*1.1
v0 = 4.69

Bent:
4.69^2 - 0^2 = 2*.5*a
a = 21.996

60*22 = F(jump) - 600
F(jump) = 1920 N
---
Thanks!
 

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