# Is there difference in force when the height increases?

Consider a man with mass 50kg.

1. When he jumps from a 2 meter height, the total force is $F=ma$ ie (mass * gravity), but nothing happens to him.

2. When he jumps from a 50 meter height, the total force is (mass * gravity ) same, but he dies.

Why is this happening?

• It's not the force exerted by gravity that kills him, it's the force exerted by the floor. Think of force as the rate of change of momentum and think about the respective momenta in each case. – Rammus Aug 19 '15 at 18:25
• You must be a young'n. I doubt I could fall 2m onto a hard surface without sustaining an injury. – Solomon Slow Aug 19 '15 at 20:08

You're looking at the net force acting on him at the instant he begins falling. And you are correct that, in the absence of air resistance, the net force that is acting on the man the instant after he begins falling in the same, given by $F_g=mg$. However, there's an old saying that says, "it's not the fall that kills you; it's the sudden stop at the end." This couldn't be more true, so to understand the difference between the two falls, we need to look at the force acting on the man as he's hitting the ground.

One good approximation to look at this is through the idea of impulse. Impulse is defined as the change in momentum of an object. At the instant before the man strikes the ground in each scenario, he has some momentum, $p$, given by $p=mv$ where $m$ is the mass of the man (in your example, 50 kg) and $v$ in the instantaneous velocity of the man right before he strikes the ground. At this point I'll do a little bit of hand-waving, and just say that the velocity right before striking the ground will increase as the height increases. Therefore, the $v$ before striking the ground for the 50m drop will be much higher than the $v$ from the 2m height. For the purpose of this problem, exact velocities are not important.

Ok, so now let's have the guy actually hit the ground. Although you may think that this occurs instantaneously, it actually occurs over a non-zero time interval. This is easily seen if you ever watch a collision through the aid of slow-motion. Anyway, so when the guy hits the ground, it takes some non-zero time interval, $\Delta t$ to come to rest. Assuming the guy is hitting the same surface in both scenarios, we'll assume $\Delta t$ is a constant across our trials. (This isn't quite true, but slight variations in $\Delta t$ across drop heights are negligible compared to wildly different impact velocities.)

When the guy hit's the ground, his momentum quickly goes from $p$, which is said is equivalent to the expression $mv$ to zero. This is because after he's fully collided with the ground, he's not moving anymore so his velocity, and therefore his momentum is zero. There's an equation that can be used to express this change in momentum, or impulse, in terms of force, and it's given by $$F\Delta t=m\Delta v$$ If the final velocity is zero, then $\Delta v$ is just given by $-v$, where $v$ is the velocity just prior to striking the ground. Solved for the impact force, $F$, we get that $$F=\frac{mv}{\Delta t}$$

Since the mass of the man, $m$, is the same no matter what height he is dropped from, and I said that we can (for our purposes) treat the time interval $\Delta t$ as a constant as well, you can see that the impact force, $F$, is proportional to the velocity $v$ the jumper has when he hits the ground. And as a mentioned earlier, higher drop height means higher final velocity, and therefore larger impact force.

When you jump from any given height, the force pulling you down is gravity with $F=mg$. This makes you accelerate to faster speeds as you fall farther, obviously.

When you hit the ground, you do not experience the same acceleration. Otherwise, it would take just as long to stop falling as it took to get up to that speed. Hitting the ground imparts a much larger acceleration. Because hard ground is practically incompressible, whatever speed you are going at when you hit ground must be lost almost immediately. To do this in such a short amount of time, you undergo a massive acceleration, and thus a massive force. This is why falling from greater height damages you more.

This is why it's a good idea not to lock your knees when landing. If you allow your knees to bend on impact, you can extend the distance and time your body has to decelerate in; much like the crumple zone on cars. However, if the acceleration required goes beyond a certain limit, the force exerted on you will shatter bones and rupture tissue no matter what. This results in the unfortunate splatting that everyone hopes to avoid.

Jim's Safety Tip: Splatting is bad.

Recall that force is equivalent to,

$$F=\frac{dp}{dt} \sim \frac{\Delta p}{\Delta t}$$

Where $p$ is the momentum, and $t$ is time. Momentum is given by,

$$p=m \cdot v$$

Where $m$ is mass, and $v$ is velocity.

When you hit the ground falling from 2 meters. The change in velocity is very small, and the change in time is small.

When you hit the ground falling from 50 meters, the change in velocity is very large, and the change in time is small.

This means the hitting force goes up, or is proportional to the initial height. Bigger fall, results in a bigger force.

The problem stems from the fact that the ground your hitting changes your velocity within the same time frame whether or not you fall 2 or 222 meters.

It's clear, higher you go, the Kinetic Energy increases. Let's see how :

$v =\sqrt{2ax}$

where $x$ here is the $h$ and $a$ here is $g$. so we can write it like that:

$v =\sqrt{2gh}$

And we have the Kinetic energy formula

$K=\frac{1}{2}mv^2$

so, as you see, increasing $h$ causes a bigger $v$ that a bigger $v$ effects on the Kinetic Energy you will have when you hit the land from height $h$.