Calculating the Force Required for an Exceptional Vertical Jump

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Homework Help Overview

The discussion revolves around calculating the force required for a vertical jump, specifically focusing on a scenario where a 70kg person jumps to a height of 0.80m after lowering themselves by 0.2m. Participants are examining the forces involved and the calculations leading to the required force.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial velocity needed for the jump and the forces acting on the jumper, including gravitational force. There are questions about the calculations leading to different force values and the interpretation of the jump's height and motion phases.

Discussion Status

Some participants are exploring different methods to calculate the necessary force, while others are questioning the assumptions made regarding the distances and forces involved. Guidance has been offered on breaking down the motion into phases and using energy conservation principles.

Contextual Notes

There is a noted discrepancy in the calculations regarding the distance covered during the jump and the forces involved. Participants are also considering the implications of gravitational force and the need to account for the initial push against the ground.

Tensaiga
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Ok here's my question:
An exceptional vertical jump from rest would raise a person 0.80m off the ground. to do this, what constant force would a 70kg person have to exert against the ground? assuming the person lowers himself by 0.2m prior to jumping and remains ina standing position while in the air.

ANS-the correct answer says in our book is 3.4 * 10^3 N. I didn't get that, i got something like 1300N.

the initial velocity i found was 3.43m/s, and i know it must over come the weight (gravtional field strength), and THAT's all i know...

Thanks
 
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Tensaiga said:
Ok here's my question:
An exceptional vertical jump from rest would raise a person 0.80m off the ground. to do this, what constant force would a 70kg person have to exert against the ground? assuming the person lowers himself by 0.2m prior to jumping and remains ina standing position while in the air.

ANS-the correct answer says in our book is 3.4 * 10^3 N. I didn't get that, i got something like 1300N.

the initial velocity i found was 3.43m/s, and i know it must over come the weight (gravtional field strength), and THAT's all i know...

Thanks
How did you get 3.43 m/s? I get something different.

How did you get 1300 N?
You must first find the acceleration of the person as he is pushing himself against the ground. what did you get?

EDIT: Btw, I get the same answer as the book.

Patrick
 
Last edited:
what...i must have done something wrong, ok i used Vf^2 - Vi^2 = 2ad, (d=0.8-0.2=0.6m), Vi = rest, 0m/s, Vf = (that's what I'm trying to find), then deceleration due to gravity, 9.8m/s^2 downwards, and you plug it in it gives you 3.43m/s, then i used that to find the time which is 0.35s (d = (v1 + v2)/2 * t), then i found out accerlation funny, it's 9.8 again, then i used to to find Fnet, and finally plus fg. and i got 1370N
 
wouldn't you want to add .8 to .2 for a total distance of 1 meter?
 
but then what would you do after that?
 
Tensaiga:
You are going down a wrong track here!

First, determine the jumper's initial velocity just when he leaves ground.
Use energy conservation for this.

Secondly, find the jumper's average acceleration from the time he has his legs lowered till he leaves ground.

Thirdly, determine the average force he's exerting on the ground.
 
Tensaiga said:
what...i must have done something wrong, ok i used Vf^2 - Vi^2 = 2ad, (d=0.8-0.2=0.6m), Vi = rest, 0m/s, Vf = (that's what I'm trying to find), then deceleration due to gravity, 9.8m/s^2 downwards, and you plug it in it gives you 3.43m/s, then i used that to find the time which is 0.35s (d = (v1 + v2)/2 * t), then i found out accerlation funny, it's 9.8 again, then i used to to find Fnet, and finally plus fg. and i got 1370N
You are going about it the wrong way, I am afraid. Arildno already explained how to do it, I will just add a few comments.

You must decompose the motion in two parts.

The first part, the person pushes against the ground with his feet. The displacement is 0.2 meter.

The second part, the person is in free fall. There the displacement is 0.8 m (not 0.6 m). Then the only force acting is the gravitational force.

Start from the second part to find the initial velocity just as the person is leaving the ground (that's using the equations you used except that the distance is 0.8 m. No need to find the time! And of course the acceleration is simply 9.8 m/s^2 downward)

Now, from knowing the final velocity when he leaves the ground, go back to part a. For that part, the initial velocity is zero, the final velocity is what you have just found in part b, you need to find the acceleration during that part (which is NOT 9.8 m/s^2 since the weight is not the only force). Once you find the acceleration, set up a free body diagram, find the normal force exterted by the ground on the person. From the action-reaction principle, this is also the force exerted by the person on the ground. You are done.

Hope this helps

Patrick
 
i see i see thank you every much
 

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