# Force of impact for human falling different heights

• brettsyoung

## Homework Statement

I am trying to prove a human head hitting the ground after falling backwards down a hill suffers a significantly greater impact than one hitting the ground on a flat surface. The person is 95KG, 180cm tall, and the impact point of the head was 60cm below (in altitude) the point where the feet were placed at the time of falling.

## The Attempt at a Solution

Unfortunately I have no physics background and have no idea where to start. This is one element of a broader forensic brief I would like to put together regarding a theoretical random assault in a sport's stadium. At some stage I understand I will need a professional report on this element of the brief, but first I would be very grateful for a ballpark assessment on whether this is indeed a relevant issue, and whether forces and hence injury are significantly greater as the slope comes into play. In other words, does pushing hooligans down the stands risk greater physical threat then pushing them up. Inituitively it does, but infact I'm less sure My schoolboy physics doesn't go beyond its ability to confuse me, so apologies if this seems a dumb question. Rather I hope it intrigues some of you. Thanks Brett

does pushing hooligans down the stands risk greater physical threat then pushing them up.
Ask yourself, "How far are they falling?"

## Homework Statement

I am trying to prove a human head hitting the ground after falling backwards down a hill suffers a significantly greater impact than one hitting the ground on a flat surface. The person is 95KG, 180cm tall, and the impact point of the head was 60cm below (in altitude) the point where the feet were placed at the time of falling.

## Homework Equations

The kinetic energy of the head is given by KE = 1/2 mv2. Rather than the head traveling 180 cm on flat ground, it travels 180 cm + 60 cm on an incline. The kinetic energy is acquired by the loss of gravitational energy U = mgh. g is 10 m/sec2 and h is either 180 cm or 240 cm. The velocity of the head is

v = √(2⋅10 m/sec2⋅.24 m) on the incline, or
v = √(2⋅10 m/sec2⋅.18 m) on flat ground

Notice that the mass does not enter into the final result.

v = √(2⋅10 m/sec2⋅.24 m) on the incline, or
v = √(2⋅10 m/sec2⋅.18 m) on flat ground
I think you mean 2.4m and 1.8m.
But is the velocity (i.e. momentum) the key measure or is it v2 (energy)?

Thanks for your replies. Very useful! Velocity is a suitable measure as we have lots of data of the effect of impact at increasing rates of velocity on the human head, including key low velocity ranges where the relative effect increases at a greater rate than the velocity. Thanks again. I didn't expect to hear back so quickly! regards Brett

I think you mean 2.4m and 1.8m.
But is the velocity (i.e. momentum) the key measure or is it v2 (energy)?
You're correct - I meant 1.8 m and 2.4 m