Jump Efficiency TheoryI am young so please no mocking.

[Karimspencer]

"Jump Efficiency Theory". I am young so please no mocking.

This is a theory i have created that judges how efficient a jump is with respect to a specific distance. This theory is going to be used mainly for the Long Jump sport. For example let's say you have to jump across a whole with a distance of 15m. And you jump with a height of 10m and a velocity of 1m/s. Now to know if your going to jump across the whole , you are going to have to calculate how efficient your jump is with respect to this distance of 15m. Now efficiency = (V*h)/d(Velocity * Height / distance) = (10*1)/15=0.666666

I came up with the equation because usually when you want succeed in jumping over a specific distance , you are going to have to jump with a very high height and with a very high speed.

Now if the efficiency of the jump is less than 1 , then your jump isn't efficient and you aren't going to pass through that specific distance but if it's higher than 1 , you will succeed.

Now remember , this is only a theory and i am only a 13-year-old boy with a passion for physics so it's probably wrong.What do you think?

 

[K^2]

That makes absolutely no sense. What you want is the range formula.

$$d=\frac{v^2}{g}sin(2\theta)$$

Here, v is your velocity, g is acceleration due to gravity, and θ is the angle of the jump. Note that optimal angle is θ=45°, at which sin(2θ)=1. Because its easier to build up horizontal speed than vertical for the long jump, the realistic angles tend to be shallower.

Suppose you wanted to jump 15m under ideal conditions. Let's take 45° angle, and g=9.8m/s² corresponding to jump on Earth. You get v=12m/s, or 27 miles per hour. That's a bit faster than a human can run. And in fact, that distance is significantly higher than a long jump world record, which is a little under 9m.

 

[mfb]

12m/s is quite close to the peak velocity of 100m-runners. If they would get some sort of ramp, they might be able to jump ~15m wide.

 

[Karimspencer]

Oh ok.

 

[haruspex]

Some thoughts you might find useful in future.
Your expression v.h/d cannot be right for a couple of reasons that should be easy to see. Dimensionally it is (L/T).L/L = L/T. That is, the net dimension is that of a speed. For an efficiency coefficient you would need it to be dimensionless.
Secondly, a large d for a given h and v would surely be highly efficient, but your expression would give it low efficiency.
Using K^2's formula, a suitable efficiency measure E would be (d.g)/(v.v), but it is more convenient for you to have this in terms of d and h.
Writing t for tan(theta) we have:
E = (d.g)/(v.v) = sin(2 theta) = 2.t/(1+t.t).
t = 2h/d.
E = 4h.d/(d.d+4h.h)