Calculating the Force Required for an Exceptional Vertical Jump

In summary, the conversation discusses the steps to calculating the constant force needed for a 70kg person to exert against the ground in order to perform an exceptional vertical jump of 0.80m. The correct answer is 3.4 * 10^3 N, which was found using energy conservation and decomposing the motion into two parts. The initial velocity was found to be 3.43m/s, and the average acceleration was determined to be 9.8 m/s^2. The conversation also includes different approaches and clarifications on how to find the correct answer.
  • #1
Tensaiga
43
0
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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  • #2
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:
  • #3
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
 
  • #4
wouldn't you want to add .8 to .2 for a total distance of 1 meter?
 
  • #5
but then what would you do after that?
 
  • #6
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.
 
  • #7
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
 
  • #8
i see i see thank you every much
 

What are Newton's three laws of motion?

Newton's three laws of motion are the principles that govern the behavior of objects in motion. The first law, also known as the law of inertia, states that an object at rest stays at rest and an object in motion stays in motion with the same speed and direction unless acted upon by an external force. The second law states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. Finally, the third law states that for every action, there is an equal and opposite reaction.

How do Newton's laws apply to jumping?

When jumping, Newton's laws can be observed in action. The first law explains why our bodies stay in motion once we jump, until the force of gravity pulls us back down. The second law explains how the force we exert on the ground to jump is directly related to the height and distance we can jump. The third law explains why we feel the ground pushing back on us as we jump, providing the necessary force to propel us upward.

What is the relationship between force and acceleration in jumping?

According to Newton's second law, the force we exert on the ground to jump is directly proportional to the acceleration we experience. This means that the more force we use to push off the ground, the greater our acceleration will be, resulting in a higher and farther jump.

How does mass affect jumping?

Newton's second law also states that the acceleration of an object is inversely proportional to its mass. This means that the more mass an object has, the less it will accelerate when the same amount of force is applied. In terms of jumping, this means that a heavier person will not be able to jump as high or far as a lighter person, even if they use the same amount of force.

Can Newton's laws be applied to other types of jumping, such as on a trampoline?

Yes, Newton's laws can be applied to any type of jumping, including on a trampoline. The force exerted on the trampoline surface to jump is similar to the force exerted on the ground, resulting in a similar acceleration and height. Additionally, the third law explains why we bounce back up after jumping on a trampoline, as the trampoline surface pushes back on us with an equal and opposite force.

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