Vertical Jump Calculations

## Question:

Please forgive me for being stupid but I can't seem to figure out that w*h=F*d stuff I understand it to a certain extent. It's just when I get a pen and pad my answer comes out different. Can you simply any more than you have to help me I'm very interested in this topic of force needed for work to be done especially vertical jumping ability.

My example: Say these person's weigh 175, 200, 225 and they all have 35 inch verticals. What would there foot pounds expended be? Say these same person's can only expend one times their own body weight what would there vertical be? What I'm really trying to figure out is if you had a bowling ball that weighed 100 lbs. and you drooped it from 12 inches, and it landed onto a catapult where you had another ball weighing 100 lbs. How high would the ball on the caterwaul go. What is the lowest amount of force an object such as a human have to expend to elevate themselves at lest 1 inch off the ground? Can the force be less then the person body weight being that if I dropped a 100 lb. bowling ball from 2 inches on a wood fall it wouldn't bounce real high.

A 175 pound person who jumped 35 inches would jump vertically 35inches/12 inches per foot or 2.917 feet. Multiply that height times the weight of 175 pounds to get the number of foot pounds of energy required to elevate a 175 pound person 2.917 feet. I get 510.417 foot pounds of energy required. Now of course once the person's feet leave the floor he can no longer contribute more energy. He will have to coast up 2.917 feet. This means that by the time he has come out of his crouch and reached full leg extension, he must have provided 510.417 foot pounds of energy beyond that which it took to just straighten him up.

Here is where we must take into account how much he elevated his center of mass from the crouched position to full leg extension. During this part of the jump he is doing two things. He is providing the energy to straighten himself out against the force of gravity and he is providing the energy that he will have to have to coast up 2.917 feet after his feet leave the floor. We need to make some assumption or measurement or in some way come up with how much he moves his center of mass in springing from the crouched position to full leg extension. Let's say we decide that he crouches enough to lower his center of mass 24 inches or 2 feet.

Now when he springs up to full leg extension, he exerts a force against the floor of some number of pounds that we want to determine. That force, whatever it is, times the two feet over which it is applied is the total number of foot pounds energy provided by the jumper. We know that that total must serve to do two things. It must get him straightened up and have enough left over to let him coast up 2.917 feet. The coasting part is 510.417 foot pounds. The straightening part is 2feet times the 175 pound weight or 350 foot pounds. The total is 860.417 foot pounds. this total is provided by the unknown force exerted over a two foot distance so if we divide the total foot pounds by 2 feet, we get the unknown number of pounds. I get 430.2085 pounds. In reality the force a person exerts against the floor in a jump is not uniform throughout the leg extension. It will be larger than average initially and drop off to zero over the last few inches of extension. In the example above we are assuming that 430.2085 pounds was the average force.

To do the other examples just plug in the other weights into the logic outlined above.

If a person could only provide enough force to equal their weight, their vertical jump would be zero. They would have all they could do to straighten up from the crouched position.

In your bowling ball example, neglecting friction, the one ball would rise to exactly the height from which the other was dropped. conservation of energy requires this.

To get the answer to your 1 inch vertical jump question, just replace the 35 inch number we used above with 1 inch.

This information is brought to you by M. Casco Associates, a company dedicated to helping humankind reach the stars through understanding how the universe works. My name is James D. Jones. If I can be of more help, please let me know. JDJ