Calculating Force for an Exceptional Standing Jump: Physics Homework Problem

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In summary, to raise a 66 kg person 0.80m off the ground, a force of 517.968 J must be exerted against the ground. This can be calculated using the formula Work = Force * Distance, assuming the person crouches a distance of 0.20m prior to jumping.
  • #1
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 66 kg person exert against the ground?
Assume the person crouches a distance of 0.20 m prior to jumping,and thus the upward force has this distance to act over before he leaves the ground.



This question i don't even know how to start
can some one explain how to even start it?
 
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  • #2


Work = Force * Distance
Force = (9.81)(66)
Work = (9.81)(66)(.80)
Work = 517.968 J

Doesn't account for the crouching distance, but it may be irrelevant, someone else will probably cover that. I hope I went in the right direction with this.
 
  • #3


Sure, I can explain how to start solving this problem.

First, we need to understand the concept of force. Force is a physical quantity that describes the interaction between two objects. In this case, the force we are interested in is the force exerted by the person on the ground, which is also known as the normal force.

Next, we need to understand the concept of work. Work is defined as the product of force and displacement. In this case, the displacement is the height the person jumps, which is 0.80m.

Using these concepts, we can set up the equation:

Work = Force x Displacement

We know the displacement (0.80m) and we need to find the force. However, we also need to take into account the crouching distance of 0.20m. This means that the person's body will already be 0.20m closer to the ground before they start jumping. So, the total displacement that the force will act over is 0.80m + 0.20m = 1.00m.

Now, we can rearrange the equation to solve for force:

Force = Work / Displacement

Plugging in the known values, we get:

Force = (66 kg x 9.8 m/s^2) / 1.00m

= 646.8 N

Therefore, the force that a 66 kg person must exert against the ground to perform an exceptional standing jump of 0.80m is 646.8 Newtons.

I hope this explanation helps you understand how to approach this problem. Remember, in physics, it's always important to understand the concepts and units involved before plugging in numbers and solving equations.
 

FAQ: Calculating Force for an Exceptional Standing Jump: Physics Homework Problem

What is the basic concept of a "jumping physics problem"?

The basic concept of a "jumping physics problem" is to determine the trajectory and motion of an object as it jumps or moves through the air, taking into account factors such as gravity, air resistance, and initial velocity.

How is the initial velocity of an object calculated in a jumping physics problem?

The initial velocity of an object is typically calculated using the equation v = √(2gh), where v is the initial velocity, g is the acceleration due to gravity, and h is the height of the object's starting position.

What role does air resistance play in a jumping physics problem?

Air resistance, also known as drag, affects the motion of an object by slowing it down as it moves through the air. This is an important factor to consider in a jumping physics problem, as it can significantly impact the object's trajectory and final landing point.

How does the mass of an object affect its jumping physics problem?

The mass of an object does not directly affect its jumping physics problem, as the acceleration due to gravity is the same for all objects. However, a heavier object will experience more air resistance, which can impact its trajectory and landing point.

Can jumping physics problems be applied to real-world scenarios?

Yes, jumping physics problems can be applied to real-world scenarios such as sports, engineering, and even everyday activities like jumping off a diving board. Understanding the principles of jumping physics can help us make more accurate predictions and improve our performance in these activities.

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