Finding the Average Force Acting on the Floor During a Vertical Jump

In summary, a student of height H jumps vertically up from the squat position. The center of mass is at a height 3h/4 from the ground, and the mass of the student is m. The free fall trajectory starts from the height of h/2 (when the student is upright and ready to lose contact) to 3h/4. The force is applied from the height of h/4 to h/2, and the work done is W = \Delta E = \Delta (KE + PE).
  • #1
lektor
56
0
Greetz :)

I recently found this problem and have made a few attempts at answering it but still have not found an answer i am 100% confident with, I would appreciate it if anyone who is keen to attempt it could give some advice and explanations for their reasoning in the solving of this question :)

A student of height H jumps vertically up from the squat postion. At the top point of the jump, the student's center of mass is at a height 3h/4 from the ground. Find the average force F acting on the floor prior to the moment when the student loses contact with the floor. It is known that the when the student stands on the floor, the center of mass is at a height h/2 from the floor; in the "squat" position, the center of mass is at a height h/4 from the floor. The mass of the student is m.
 
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  • #2
Some initial thoughts:

The free fall trajectory starts from the height of h/2 (when the student is upright and ready to lose contact) to 3h/4. So, in effect the CM climbs by h/4 till the height becomes zero. So you can find the initial speed and KE = 1/2Mv^2.
The force is applied from the height of h/4 to h/2, so it is applied over a distance of h/4. Use the Work-Energy theorem:

[tex]W = \Delta E = \Delta (KE + PE)[/tex]

If you find work, then it's easy:

[tex]F_{av} = \frac{W}{\Delta h}[/tex]
 
  • #3
Yeah i see what you mean and this was one of my methods but it didn't achieve a final answer.. or a definative forumla..
 
  • #4
lektor said:
Yeah i see what you mean and this was one of my methods but it didn't achieve a final answer.. or a definative forumla..

I don't see any point where you get stuck. It's simple!
Just to clarify something else, the work W is the sum of the positive work done by the muscles and the negative work done by (constant) gravity. So you need to specifically determine the work done by the legs of the leaping person.
 
  • #5
ramollari said:
I don't see any point where you get stuck. It's simple!
Just to clarify something else, the work W is the sum of the positive work done by the muscles and the negative work done by (constant) gravity. So you need to specifically determine the work done by the legs of the leaping person.


lol oh my...
sorry ! got it now thankyou ! :D
 

Related to Finding the Average Force Acting on the Floor During a Vertical Jump

1. What is the center of mass and why is it important?

The center of mass is a point in an object or system where the mass is evenly distributed in all directions. It is important because it helps determine how an object will move and behave under the influence of external forces.

2. How is the center of mass calculated?

The center of mass can be calculated by finding the weighted average of the positions of all the individual particles that make up an object.

3. Can the center of mass be outside of an object?

Yes, the center of mass can be outside of an object if the object has an irregular shape or if the mass is unevenly distributed.

4. How does the center of mass affect rotational motion?

The center of mass plays a crucial role in rotational motion as it determines the axis around which an object will rotate and the amount of torque needed to cause rotation.

5. How does the location of the center of mass change when objects are combined or separated?

The location of the center of mass changes when objects are combined or separated based on the distribution of mass in the new system. If the objects are combined, the center of mass will shift towards the heavier object, and if they are separated, the center of mass will shift towards the lighter object.

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