Ground reaction forces during a vertical jump
Ground reaction forces are the forces that the ground exerts on a jumper during the course of a vertical jump.

You’re maybe asking yourself: “Isn’t it the other way around? Isn’t the athlete exerting forces on the ground instead of the ground on the athlete?”

You are not wrong! In fact, both forces occur during a vertical jump. This phenomenon is explained in the 3rd Newton Law:


To every action, there is always opposed an equal reaction.
Newtons 3rd Law
So if an athlete is exerting forces on the ground, then the ground is always exerting the same force, in the opposite direction on the athlete.

Force Plate Analysis
Sports scientist are able to measure these ground reaction forces with a technology called force plates. These plates record the exact forces occurring during a vertical jump (or any other movement) and allow you to see how quickly athletes can produce forces, how large these forces are, and to expose potential imbalances between the left and right leg.

For an example of a force plate analysis check out this Youtube video:

Slow motion video of vertical jump with synchronized vertical force data.
Analyzing the forces of a vertical jump
In the following paragraphs, we want to look at the relations between forces, acceleration, speed, and height of a vertical jump. For that, we are going to use this example of a force plate analysis. It is a little simplified. as in reality the force curve wouldn’t be as smooth, but it will work just fine as an example:

Graph 1: Ground reaction forces during vertical jump:

Force of Jumper Movement

Force of Gravity
Phase 1: Before the Jump
Before the jump, the analysis shows a flat line at a force of 981 Newton. The athlete is not moving at that time, so where are these forces coming from?

Of course, this is gravity which is pulling the athlete toward the ground. This force is explained in this formula:

\displaystyle \ F = m * g
Where m is the mass of the athlete and g is the acceleration of earth’s gravity.

F is the amount of force the athlete has to exert at any moment just to neutralize the forces of gravity.

We know that g=9.81 {m}/{s^2} on earth, therefore:

\displaystyle \begin{aligned} 981 N &= m * 9.81 m/s^2 \\ \\ => m &= \frac{981N}{9.81 m/s^2} = 100kg \end{aligned}
So, before the athlete starts any motion, the force plate basically acts as a simple weighing scale, showing the amount of force that gravity exerts on the athlete.

Phase 2: Descending into the Jump
1. Accelerating the downward movement:
During 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 resist gravity, which means the athlete is accelerating a downward movement.

The acting forces at this time can be described as follows:

F_{Jumper} = F_{GRF} - F_{Gravity} <=0
One measure that is interesting to sports scientist is how fast the athlete is moving downwards before the jump. But are we able to calculate that from the recorded forces of the analysis?

We know that:

F = m a => F t = m a t
Because F is not constant but a function of time, and v =a t :

\displaystyle => \int_{t_1}^{t_2} F_{Jumper}(t) \mathrm{d}t = m v
where F_{Jumper}(t) is the difference between the registered ground reaction forces and gravity.

This integral can be numerically calculated from the data obtained by the force plate analysis. The downward impulse generated by the jumper is shown in the graph as the red area below the line representing gravity:

Graph 2: Forces during Phase 2

Force of Jumper Movement

Force of Gravity
Let’s assume a numerical estimation finds the integral (impulse) to be -70 Ns. We can then conclude that:

\displaystyle \int_{t_1}^{t_2} F(t) \mathrm{d}t = m v \\ \\ \\ => -70 N s = 100kg * v \\ \\ \\ => v = -70 N s / 100kg = -0.7 m/s
Therefore, the athlete reaches the highest velocity of 0.7 meters/second during the downward movement preceding the jump.

2. Decelerating the downward movement
So far we only looked at the first part of the countermovement, where the athlete is accelerating his downward movement. This part of the movement can be easily identified on the force-time diagram as the part where ground reaction forces are below the expected forces due to gravity.

The second part of the countermovement is less obvious. The athlete has to decelerate the downward movement to reach a short moment of pause at the deepest position of the jump.

Can we find this moment in the graph of the force plate analysis?

We have seen that during the first part of phase two we accumulated a downward impulse of 70Ns which resulted in a velocity of 0.7 m/s

We are now looking for an equally big impulse in the opposite direction. This impulse can be described as:

\displaystyle \int_{t_2}^{t_3} F_{Jumper}(t) \mathrm{d}t = 70 Ns
As F(t) and t_2 are known, the numerical algorithm of the force plate analysis is now looking for t_3 so that the impulse equals 70Ns.

To better picture it, imagine looking for a t_3 so that the pink and blue areas are exactly the same size:

Graph 3: Forces during Deceleration of Phase 2

Force of Jumper Movement

Force of Gravity
Phase 3: Upward Motion before takeoff
This phase begins with the athlete at the bottom of the jump, just as he begins exploding upwards towards the takeoff. The force-time graph shows that the athlete reaches peak forces shortly after reaching the lowest point of the jump. He then further accelerates until his feet leave the ground and there are no more ground reaction forces measurable.

If you want to assess the velocity during takeoff we can use the same technique as during phase 2:

\displaystyle \int_{t_3}^{t_4} F_{Jumper}(t) \mathrm{d}t = m v
This integral can be pictured as the yellow area (subtracting the small brown area right before the takeoff) in the following force-time graph:

Graph 4: Forces during Upward Motion

Force of Jumper Movement

Force of Gravity
The numerical algorithm of the force plate analysis calculates an impulse of 245Ns, therefore we can determine the initial vertical velocity during takeoff as:

\displaystyle 245 N s = 100kg * v \\ \\ \\ => v = 245 N s / 100kg = 2.45 m/s
Phase 4: The Flight
During this phase, the athlete can’t impact the velocity of his center of gravity any further. The height of the jump has been predetermined by the build up of speed before and during takeoff. The only force that is now acting upon the athlete is the gravity that is pulling the jumper back down.

If the athlete can’t do anything at this point to increase his vertical jump, can we then determine the jump height using the recorded ground reaction forces during phases 1-3?

We do know so far that the initial velocity v(0)=2.45 m/s^2 and that the gravity of earth has an acceleration of a=9.81m/s^2
We also know that during the peak of the jump, vertical velocity has to be zero as otherwise, the athlete would still gain height, which would also mean that he hasn’t reached the peak of the jump yet.

\displaystyle v(t_{peak})=0
If we know initial velocity and earth’s gravity, we can calculate the velocity during every moment of the jump like this:

\displaystyle v(t)=v(0) -a * t
Therefore, we can calculate the time it takes the jumper to reach the peak of the jump as follows:

\displaystyle \begin{aligned} v(t_{peak}) &= v(0) - a * t_{peak} \\ \\ => 0 &= v(0) - a * t_{peak} \end{aligned}
\displaystyle = > t_{peak} = \frac{v(0)}{a} = \frac{2.45m/s}{9.81m/s^2} = 0.25s
So now that we know the velocity v(t) during every moment of the jump, and the time of the peak of the jump 0.25s we can calculate the jump height as the integral of velocity over the time it takes to reach the peak of the jump:

\displaystyle \begin{aligned} h_{jump} &= \int_0^{\frac{v(o)}{a}} \bigg( v(o) - at \bigg) \ \mathrm{d}t= \\ \\ & = v(o)t \

Totally not copied from https://www.thehoopsgeek.com/the-physics-of-the-vertical-jump/ !!! I would never!!!!!!!!!!!!!!