Calculating Force Required to Perform an Exceptional Standing Jump

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Homework Help Overview

The discussion revolves around calculating the force required for a 66 kg person to perform an exceptional standing jump that reaches a height of 0.80 m. The problem involves understanding the forces at play during the jump, particularly the force exerted against the ground, and considers the crouching distance of 0.20 m prior to the jump.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between force, work, and energy in the context of the jump. Some discuss using kinematics to find the initial velocity needed to reach the jump height, while others question how to derive acceleration from the given parameters. There are attempts to equate work done with potential energy and discussions about the equations of motion relevant to the problem.

Discussion Status

The conversation is ongoing, with participants sharing various approaches and equations. Some have suggested using kinematics exclusively, while others have pointed out the importance of understanding work and energy concepts. There is a recognition of the complexity of the problem, and participants are actively questioning and clarifying their understanding of the relevant physics principles.

Contextual Notes

Participants note that they have not yet covered the concept of work in their studies, which may limit their ability to fully engage with the problem. There is also mention of needing to split the problem into two parts: the crouch phase and the jump phase, indicating a methodical approach to tackling the question.

PearlyD
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Homework Statement


An Exceptional standing jump would raise a person 0.80m off the ground. To do this, what force must a 66Kg person exert against the ground? Assume the person crouches a distance of 0.20m prior to jumping, and thus the upward force has this distance to act over before he leaves the ground.



Homework Equations





The Attempt at a Solution


Heres what i attempted.
The upward force would be the normal force(Fn) which is the same as the Force of gravity so i caculated that and i got Fg=646.8 the force going down would be the applied force(Fa)
Heres what i got fo far before i got stumped..
Fa-Fn=ma
Fa-646.8=66a

I don't know how i would get acceleration?
 
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PearlyD said:

Homework Statement


An Exceptional standing jump would raise a person 0.80m off the ground. To do this, what force must a 66Kg person exert against the ground? Assume the person crouches a distance of 0.20m prior to jumping, and thus the upward force has this distance to act over before he leaves the ground.



Homework Equations





The Attempt at a Solution


Heres what i attempted.
The upward force would be the normal force(Fn) which is the same as the Force of gravity so i caculated that and i got Fg=646.8 the force going down would be the applied force(Fa)
Heres what i got fo far before i got stumped..
Fa-Fn=ma
Fa-646.8=66a

I don't know how i would get acceleration?

First, The force of the jump must provide an acceleration greater than gravity or he will never leave the ground. One of the more straightforward approaches is that we can equate the work done to the force * displacement (here given as 0.2 meters) with the potential energy given which is the displacement of the center of mass (0.8m) x gravity*mass.

Since you know the force, the acceleration as he straightens his legs can be found.
 
denverdoc said:
First, The force of the jump must provide an acceleration greater than gravity or he will never leave the ground. One of the more straightforward approaches is that we can equate the work done to the force * displacement (here given as 0.2 meters) with the potential energy given which is the displacement of the center of mass (0.8m) x gravity*mass.

Since you know the force, the acceleration as he straightens his legs can be found.

What do you mean by work done?
Also do you mean by this equation..
Work done* Displacement(0.2) EQUALS potential energy?
and Potential energy is also 0.80*Gravity*mass?
 
Have you covered work? If not we can use a purely kinematics approach. But by work yes the upward force times the displacement over which it acts--as he goes from a crouch to the point at where his feet leave the ground. After that he can exert no more force and he slows due to gravity.

EDIT: But yes, you have almost the right eqns listed. Work=force* displacement=ma* displacement
 
denverdoc said:
Have you covered work? If not we can use a purely kinematics approach. But by work yes the upward force times the displacement over which it acts--as he goes from a crouch to the point at where his feet leave the ground. After that he can exert no more force and he slows due to gravity.

EDIT: But yes, you have almost the right eqns listed. Work=force* displacement=ma* displacement

No, i haven't covered work.
So from crouch to the point where his feet leave the ground i got a force of 129.36N
How would you find the acceleration,to find the Force that he exerts againts the ground?
the only equations i know of at the moment is Fnet=ma and Fg=mg
 
Ok, forget work and potential energy.

Lets use kinematics only.

Can you calculate the initial velocity of an object leaving the ground which reaches 0.8meters high?
 
denverdoc said:
Ok, forget work and potential energy.

Lets use kinematics only.

Can you calculate the initial velocity of an object leaving the ground which reaches 0.8meters high?
To find an initial velocity id either need a Final velocity or a time. and i don't have either.
 
No you don't: this is the equation I used in the similar problem which in completeness is

2*a*(displacement)=Vf^2-Vi^2 In both these problems, either initial or final velocity is zero.


You can derive these equations by calculating time, but they are of such usefulness in kinematics, well worth committing to memory. Another tip is that in these types of problems you could have considered the mass at rest at 0.8m hiigh and ask yourself how fast it hits the ground--it is exacty the same problem as asking how fast must it be going leaving the ground to reach a height of 0.8 m.
 
denverdoc said:
No you don't: this is the equation I used in the similar problem which in completeness is

2*a*(displacement)=Vf^2-Vi^2 In both these problems, either initial or final velocity is zero.


You can derive these equations by calculating time, but they are of such usefulness in kinematics, well worth committing to memory. Another tip is that in these types of problems you could have considered the mass at rest at 0.8m hiigh and ask yourself how fast it hits the ground--it is exacty the same problem as asking how fast must it be going leaving the ground to reach a height of 0.8 m.

Is 2a the same as displacement?
 
  • #10
PearlyD said:
Is 2a the same as displacement?


No the displacement is the distance over which the force acts: the a is acceleration.

This will allow you to get acceleration and hence force like the other problem.
 
  • #11
denverdoc said:
No the displacement is the distance over which the force acts: the a is acceleration.

This will allow you to get acceleration and hence force like the other problem.

I used (mass)Gravity)(Displacement)=1/2mV^2
to get a velocity,would that work for this question?
 
  • #12
That works to get the initial velocity. But tere are two displacements in the problem--the first is that of the center of mass as he springs from crouch to leap. The other displacement is that against gravity--how high he flies.

(This is why concept of work would be so useful: force1*displacement1 is equal and opposite to force2 * displacement2). The mass is the same in both cases. So the relative accelerations are proportional to the distance over which the forces act)
 
  • #13
denverdoc said:
That works to get the initial velocity. But tere are two displacements in the problem--the first is that of the center of mass as he springs from crouch to leap. The other displacement is that against gravity--how high he flies.

(This is why concept of work would be so useful: force1*displacement1 is equal and opposite to force2 * displacement2). The mass is the same in both cases. So the relative accelerations are proportional to the distance over which the forces act)
Im starting over again
and I am spliting the question in two
first one is the crouch,and I am going to find that force,that he uses to go from crouch to stand.
But what kinematics equation would i use to find acceleration?
I have Vi which is 0 and distance= 0.20 but that's it i don't know what id do?
 

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