When I started to work-related on the video tool that steps vertical jump, I had actually to dust off my old textbooks to learn around the connection between hang time and also jump elevation. And to my surpincrease, it turned out that the vertical jump is a great (and also interesting!) example of the regulations of physics at work. You can really learn around the relationship between velocity, acceleration, pressures and hang time. Definitely more exciting than the average instance of your physics textbook!
In this article, I am going to look at the five phases of the vertical jump and show you how physics work-related in the time of each of them. This will be accompanied by charts and also interactive calculators, and in the end, we are going to answer some fun inquiries like: “How high would Micheal Jordan jump on the Moon?”.
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Table of Contents
Questions answered by physics
A good means to look at the physics of the vertical jump is to look at the forces that occur throughout each phase:
Ground reactivity pressures throughout a vertical jump
Ground reaction forces are the forces that the ground exerts on a jumper throughout the course of a vertical jump.
You’re maybe asking yourself: “Isn’t it the other means around? Isn’t the athlete exerting pressures on the ground rather of the ground on the athlete?”
You are not wrong! In truth, both pressures take place during a vertical jump. This phenomenon is explained in the third Newton Law:
To eexceptionally action, tbelow is constantly opposed an equal reactivity.Newloads 3rd Law
So if an athlete is exerting forces on the ground, then the ground is constantly exerting the very same pressure, in the opposite direction on the athlete.
Force Plate Analysis
Sports scientist are able to measure these ground reaction forces through an innovation called force plates. These plates document the precise forces occurring during a vertical jump (or any kind of other movement) and permit you to watch how conveniently athletes can develop pressures, exactly how large these forces are, and also to reveal potential inequalities in between the left and also best leg.
For an example of a force plate analysis check out this Youtube video:
Slow activity video of vertical jump through synchronized vertical force information.
Evaluating the forces of a vertical jump
In the adhering to paragraphs, we want to look at the relationships in between forces, acceleration, speed, and elevation of a vertical jump. For that, we are going to use this example of a force plate evaluation. It is a small simplified. as in reality the force curve wouldn’t be as smooth, yet it will certainly job-related just fine as an example:Graph 1: Ground reaction forces during vertical jump:
Phase 1: Before the Jump
Before the jump, the analysis reflects a level line at a force of 981 Newton. The athlete is not relocating at that time, so wbelow are these pressures coming from?
Of course, this is gravity which is pulling the athlete toward the ground. This pressure is explained in this formula:
$latexdisplaystyle F = m * g&s=2$
Where $latex m &s=1 $ is the mass of the athlete and also $latex g &s=1 $ is the acceleration of earth’s gravity.
F is the amount of pressure the athlete hregarding exert at any kind of minute just to neutralize the forces of gravity.
We know that $latex g=9.81 m/s^2 &s=1 $ on earth, therefore:
$latexdisplaystyleeginaligned981 N &= m * 9.81 m/s^2 \ \=> m &= frac981N9.81 m/s^2 = 100kgendaligned&s=2 $
So, prior to the athlete starts any kind of movement, the force plate basically acts as a simple weighing scale, reflecting the amount of pressure that gravity exerts on the athlete.Phase 2: Descending right into the Jump1. Accelerating the downward movement:
Throughout this phase, the athlete bends his knees, swings his arms and prepares the jump by lowering the center of gravity. The force plate is registering forces lower than the 981N needed to withstand gravity, which suggests the athlete is increasing a downward motion.
The acting pressures at this time deserve to be defined as follows:
$latex F_Jumper = F_GRF – F_Gravity F = m a => F t = m a t&s=2$
Because F is not constant yet a role of time, and $latex v =a t $ :
$latex displaystyle=> int_t_1^t_2 F_Jumper(t) mathrmdt = m v &s=2 $
where $latex F_Jumper(t) &s=1$ is the distinction in between the registered ground reaction forces and also gravity.
This integral have the right to be numerically calculated from the information acquired by the pressure plate evaluation. The downward impulse produced by the jumper is shown in the graph as the red area below the line representing gravity:Graph 2: Forces during Phase 2
Let’s assume a numerical estimation finds the integral (impulse) to be -70 Ns. We can then conclude that:
$latexdisplaystyleint_t_1^t_2 F(t) mathrmdt = m v \ \ \=> -70 N s = 100kg * v \ \ \=> v = -70 N s / 100kg = -0.7 m/s&s=2 $
Therefore, the athlete reaches the highest velocity of 0.7 meters/second in the time of the downward motion preceding the jump.2. Decelerating the downward movement
So far we just looked at the initially part of the countermovement, wright here the athlete is speeding up his downward movement. This part of the movement have the right to be conveniently figured out on the force-time diagram as the part wbelow ground reactivity forces are listed below the expected forces because of gravity.
The second component of the countermotion is less evident. The athlete has to deceleprice the downward movement to reach a short moment of pause at the deepest place of the jump.
Can we find this moment in the graph of the force plate analysis?
We have viewed that during the initially component of phase two we gathered a downward impulse of 70Ns which brought about a velocity of 0.7 m/s
We are now trying to find an equally big impulse in the oppowebsite direction. This impulse have the right to be defined as:
$latexdisplaystyleint_t_2^t_3 F_Jumper(t) mathrmdt = 70 Ns&s=2$
As $latex F(t) &s=1$ and $latex t_2 &s=1 $ are well-known, the numerical algorithm of the pressure plate analysis is currently searching for $latex t_3 &s=1 $ so that the impulse equals 70Ns.
To much better photo it, imagine trying to find a $latex t_3 &s=1 $ so that the pink and blue areas are exactly the same size:Graph 3: Forces during Deceleration of Phase 2
Phase 3: Upward Motion prior to takeoff
This phase begins through the athlete at the bottom of the jump, simply as he begins exploding upwards in the direction of the takeoff. The force-time graph mirrors that the athlete reaches optimal forces soon after reaching the lowest point of the jump. He then better increases until his feet leave the ground and also tbelow are no even more ground reactivity forces measurable.
If you want to assess the velocity during takeoff we deserve to use the very same method as during phase 2:
$latex displaystyleint_t_3^t_4 F_Jumper(t) mathrmdt = m v &s=2 $
This integral have the right to be pictured as the yellow location (subtracting the small brvery own area best prior to the takeoff) in the following force-time graph:Graph 4: Forces during Upward Motion
The numerical algorithm of the force plate analysis calculates an impulse of 245Ns, therefore we deserve to recognize the initial vertical velocity during takeoff as:
$latexdisplaystyle245 N s = 100kg * v \ \ \=> v = 245 N s / 100kg = 2.45 m/s&s=2$Phase 4: The Flight
Throughout this phase, the athlete can’t impact the velocity of his facility of gravity any type of additionally. The height of the jump has been preestablished by the build up of rate before and throughout takeoff. The only pressure that is now acting upon the athlete is the gravity that is pulling the jumper back down.
If the athlete can’t carry out anypoint at this suggest to boost his vertical jump, can we then recognize the jump height making use of the videotaped ground reaction pressures throughout phases 1-3?
We do understand so much that the initial velocity $latex v(0)=2.45 m/s^2 &s=1$ and also that the gravity of earth has an acceleration of $latex a=9.81m/s^2 &s=1$
We also recognize that during the peak of the jump, vertical velocity hregarding be zero as otherwise, the athlete would certainly still acquire height, which would certainly also suppose that he hasn’t reached the optimal of the jump yet.
If we understand initial velocity and earth’s gravity, we deserve to calculate the velocity throughout eincredibly minute of the jump prefer this:
$latexdisplaystylev(t)=v(0) -a * t &s=2$
Therefore, we can calculate the time it takes the jumper to reach the height of the jump as follows:
$latexdisplaystyleeginalignedv(t_peak) &= v(0) – a * t_peak \ \=> 0 &= v(0) – a * t_peakendaligned &s=2 $
$latexdisplaystyle= > t_peak = fracv(0)a = frac2.45m/s9.81m/s^2 = 0.25s&s=2 $
So currently that we understand the velocity $latex v(t) &s=1 $ throughout every minute of the jump, and the moment of the top of the jump 0.25s we can calculate the jump elevation as the integral of velocity over the time it takes to reach the peak of the jump:
$latexdisplaystyleeginalignedh_jump &= int_0^fracv(o)a igg( v(o) – at igg) mathrmdt= \ \& = v(o)t – frac12 at^2 igg|_0^fracv(0)a \ \& = v(0) left(fracv(0)a ight) – frac12 a left(fracv(0)a ight)^2 \ \& = fracv(0)^2a – frac12 fracv(0)^2a \ \ endaligned &s=2 $
$latexdisplaystyle=> h_jump = frac12 fracv(0)^2a&s=2 $
So, we end up with a pretty basic formula that enables us to calculate the vertical jump height if we recognize the initial velocity. In our certain case we get:
$latexdisplaystyleh_jump = frac12 frac(2.45m/s)^29.81 m/s^2 = 0.306m&s=2 $
This interenergetic graph shows the connection between impulse, mass of the jumper, initial velocity and jump height:Calculate Vertical Jump Height based upon impulse, velocity and also weight:
If you are a sports scientist and you have accessibility to a pressure plate then this is great! But are tbelow less complicated ways to calculate vertical jump height using physics?Calculating vertical jump elevation from hang time
Let’s assume we have no concept about the forces during takeoff or the initial velocity of the jumper. We know, yet, the moment the jumper remained in the air (hang time). This is not a very unrealistic expectation as you deserve to meacertain hang time making use of high structure rate camages or also modern-day cell phones pretty quickly.
In our example we have measured a hang time of $latex t=0.5 &s=1 $.
We currently established that we deserve to calculate vertical jump elevation as a product of initial velocity:
$latexdisplaystyle=> h_jump = frac12 fracv(0)^2a&s=2 $
So, now we simply need to find the initial velocity for a vertical jump that takes 0.5s!
If a jumper jumps 1m high it additionally implies he hregarding fall 1m after he reaches the peak of the jump. And as velocity is a straight feature $latex (v=a t) &s=1 $ we have the right to present that the jumper reaches the optimal of the jump constantly specifically in the middle of a jump. Therefore:
$latexdisplaystylet_peak = 0.5 * t_hangtime&s=2 $
So, if we desire to recognize how high someone via a hang time of 0.5s jumped, we can just calculate the distance a complimentary falling object travels in 0.25s! In general:
$latexdisplaystyleeginalignedS &= int_0^frac12 t_hangtime v(t) ; dt \ \&= int_0^frac12 t_hangtime a * t ; dt \ \&= frac12 a t^2 ; igg|_0^frac12t_hangtime \ \&= frac12 a left( frac12 t_hangtime ight)^2 \ \&= frac18 a : t_hangtime^2 \ \endaligned&s=2 $
In our certain instance with a hang time of 0.5s we get:
$latex displaystyleS = frac18 a : t_hangtime^2 = frac18 * 9.81m/s^2 * 0.5^2 = 0.306m&s=2 $
This formula is at the heart of the vertical jump measurement app I have actually produced. If you have actually video of your jump you deserve to pack it in the internet browser, tag the takeoff and also landing, and also the application will certainly tell you your vertical jump elevation. You deserve to likewise inspect out the iPhone variation on the App store right here.
Phase 5: Landing
Throughout the takeoff an athlete generates forces that eventually lead to a vertical velocity high sufficient to leave the ground. We have actually shown before, that this vertical velocity reaches 0 at the height of the jump, and also it is easy to show that the velocity is exactly the same throughout landing as it was during takeoff (however directed in the opposite direction).
If the athlete wants to soptimal the downward activity and also concerned a standstill he hregarding exert sufficient pressures to cancel out the impulse of the vertical rate of the landing:
$latexdisplaystyleint_t_4^t_5 F(t) mathrmdt = m v(t_4) = m v(t_0)&s=2$
This impulse deserve to be pictured as the green area in the adhering to graph:Graph 5: Forces during Landing
In our instance, the athlete increased from a velocity $latex v(t_0) = 0 &s=1$ to $latex v(t_3) = 2.45 m/s^2 &s=1$, throughout the landing he hregarding accelerate from a velocity of $latex v(t_4) = -2.45m/s^2 &s=1$ to a rate of $latex v(t_5)=0m/s^2 &s=1$.
As the readjust in speed $latex Delta v =2.45m/s^2 &s=1$ is the same in both instances, the Impulse hregarding be the same as well. This can be viewed in the graph as both the yellow and also the green area result in 245Ns.
The athlete is currently ago in a standstill and also the 5 phases of the vertical jump are complete!
Questions answered by physics
Now that we have learned the partnership in between physics and the vertical jump. Let’s answer some fun questions! If you want to attempt your knowledge and also answer them yourself, you have the right to examine out the equation arsenal or the interenergetic calculator at the end of this short article.
How high would certainly Micheal Jordan jump on the Moon?
During his prime, Micheal Jordan had a vertical of at least 110cm. So much we have taken the acceleration of earth $latex a=9.81m/2^2 &s=1$ for granted. But what happens if you look at the gravity of the moon?
Answer:If we assume that MJ is able to generate the same impulse during takeoff on Moon as on Earth then his initial vertical velocity $latex v(0) &s=1$ would be unreadjusted. However, velocity during the time of trip would adjust dramatically, as it is no much longer decelerated by the gravity of earth ($latex a= 9,81 m/s^2 &s=1$) but by the gravity of Moon ($latex a = 1.622 m/s^2 &s=1$).
If we desire to get the precise height of the jump we just have to look at this formula for vertical jump height:
$latexdisplaystyleeginalignedh_jump &= frac12 * fracv(0)^2a \ \=> h_Moon &= frac12 * fracv(0)^21.622 \ \h_Moon &= frac12 * fracv(0)^29.81m/s^2 * frac9.81m/s^21.622m/s^2 \ \h_Moon &= h_Earth * frac9.81m/s^21.622m/s^2 \ \&= 110cm * 6.05 = 665 cmendaligned&s=2 $
Gravity on Moon is about 1/6th of the gravity on Earth, therefore MJ would be able to jump 6 times as high (665cm)!
Note: The assumption that MJ would have the ability to generate the very same impulse on the Moon would certainly not organize true in truth. The reduced gravity would certainly make the countermotion before the jump a lot sreduced, therefore it would be a lot harder to generate pressures as easily as on earth.
Do Jumpers really stop in the air?
Throughout NBA telecasts you regularly hear the announcers say that a player quit in the air throughout a specifically athletic dunk or block. But have the right to you really speak in the air?
Answer:The elevation of a jumper in the time of each moment of the jump is explained by this formula:
$latexdisplaystyleht = v(0)*t – frac12 a t^2&s=2$
In this interactive graph reflects the plane of flight:
We deserve to watch that a jumper spends 10% of his hang time in the peak 1% (or 50% of his hangtime in the top 25%). In general:
$latexdisplaystylepercent_hangtime= sqrtpercent_height&s=2 $
So the answer is no, the jumper is not really hanging in the air, but it shows up this means bereason the athlete spends a disproportional amount of the hang time at the peak of the jump.
See more: Flight History For American Airlines Flight 1908 (Las Vegas To Dallas
If you shed weight, will certainly you be able to jump higher?
Assuming you lose weight yet save all the strength and power, will you be able to jump higher?
If an athlete loses 10% of his body weight however continues to be able to geneprice the very same forces in the time of takeoff, then he significantly improves his takeoff velocity!
My initially idea around exactly how to calculate the adjust in jump height, wregarding assume that the impulse remains the same if the jumper loses weight yet the pressures remajor the exact same. But this means I gained results that just didn’t seem right…
It took me a while however I lastly known why! Look at the meaning of the impulse:
$latexdisplaystyleI = int_start^takeoff F(t) dt&s=2 $
$latex F(t) &s=1 $ stays the very same, but for a lighter athlete, these exact same pressures bring about a larger acceleration which in transforms provides the takeoff much faster. This suggests, that the time during which the jumper can exert pressures becomes much less and also therefore the impulse decreases as well.
A better means to look at this problem is to assume that the potential energy of the body remains constant:
Let’s take a look at the formula for the gravitational potential energy that the jumper possesses at the peak of the jump:
$latexdisplaystyleU = m * g * h&s=2 $
If we asume $latex U &s=1 $ to be constant then we get for the brand-new jump height:
$latexdisplaystyleh_old = fracUmg \ \eginaligned => h_new &= fracU0.9*mg \ \&= fracUmg * frac10.9 \ \&= h_old * 1.111 endaligned&s=2 $
So if you lose 10% bodyweight while every little thing else continues to be the very same, you would certainly improve your vertical jump by 11.1% ! If $latex x &s=1$ is the weight loss in %, then in general:
$latexdisplaystyleh_new = h_old * ( frac11-x)&s=2 $
Note: In reality, the innovation would certainly be less excessive, as losing weight virtually always additionally indicates shedding power. Also, the countermotion, in the start, would have less influence as the same downward rate would certainly result in a lower impulse as a result of the reduced body weight