Calculating Speed for High Jump of 1.60m

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SUMMARY

The required takeoff speed for a 70-kg high jumper to reach a height of 1.60 meters is 5.6 m/s. This calculation utilizes the principles of conservation of energy, specifically the relationship between kinetic energy and gravitational potential energy. The equations applied include Kinetic Energy = 0.5(mass)(velocity^2) and Gravitational Potential Energy = (mass)(acceleration due to gravity)(height). By substituting the values into the energy conservation equation, the solution confirms that the jumper must achieve a speed of 5.6 m/s at takeoff.

PREREQUISITES
  • Understanding of Kinetic Energy and Gravitational Potential Energy
  • Basic knowledge of the conservation of energy principle
  • Familiarity with algebraic manipulation of equations
  • Concept of acceleration due to gravity (9.8 m/s²)
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  • Explore the effects of mass and height on potential energy
  • Learn about the relationship between speed and kinetic energy
  • Investigate real-world applications of energy conservation in sports science
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Homework Statement


At what speed would a 70-kg high jumper have to leave the ground in order to raise his body up to a height of 1.60 m

Acceleration due to gravity: 9.8 m/s^2 [down]

Homework Equations



Kinetic Energy = 0.5(mass)(velocity^2)
Gravitational Potential Energy = (mass)(acceleration due to gravity)(height)

The Attempt at a Solution

 
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Answer is 5.6 m/s but i don't know how to get it
 
Because only conservative forces are acting upon the system:
Change in Potential Energy + Change in Kinetic Energy = 0
Change in potential energy = mg(hf-hi)
Change in kinetic energy = .5(m)(vf2-vI2)
vf is zero because the jumper just reaches the maximum height of 1.6 meters. hi is also zero because the jumper starts on the ground. hf is 1.6 meters. Substituting, we obtain
-1097.6 + 35vi2 = 0 (Because gravity points down, it has a value of -9.8 m/s20
Solving for v results in the answer of 5.6 m/s.
 

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