I fell 30 feet out of a tree. What force did I hit the ground with?

Question

I weigh about 165lbs and had say 20lbs of climbing gear on. Falling at 33ft/sec/sec I was in the air for just under one second. Could anyone tell me the force of impact? I landed somewhat flat on my back so the majour impact was spread over about say 4.5 to 5 sq ft.
(I was able to walk out of the hospital that night but spent the next 6 weeks onthe couch)

Answer 1

Ouch!

[Edit -- snoboarderk26 is calculating your WEIGHT in Netwtons, not your impact force. mavis b is pulling formulas out of thin air. Height divided by gravity does not give you your time of fall; and dividing your momentum by your time of fall does not give you your impact force.]

The total force of impact actually does not depend on how it was spread out over your body (although it's still a good idea to maximize that area, because that way you minimize the force _per_ square inch).

Knowing the height from which you fell, we can say that the _speed_ at impact was about 44 feet per second (using the formula: v = sqrt(2gh)).

However, the _force_ at impact depends critically on how long it took your body to decelerate from 44 fps to zero, once you contacted the ground. Specifically,

Force = mv/t

where m is your mass (in your case, about 84 kg); v is your speed (44 fps), and t is the deceleration time.

Some examples:
if t=1.0 second, Force = 253 lbs. (0.37 lbs per sq. in., assuming spread out over 4.75 sq. feet of body surface)
If t=0.5 seconds, Force = 506 lbs. (0.74 lbs per sq. in.)
if t=0.1 seconds, Force = 2532 lbs. (3.7 lbs per sq. in.)

This is exactly why soft things are more fun to land on than hard things--as they "give," they increase the time it takes for you to decelerate. As you can see above, even a modest increase in that time, can greatly decrease the force.

The formula can also be expressed in terms of the _distance_ over which you decelerated, instead of the time. In that case, the formula is:

Force = mv²/(2d),
where "d" is the deceleration distance.

If we assume that your center of gravity smooshed down about 3 inches as you were slamming to a stop, then the force was 22,286 pounds! (or, about 32.6 lbs. per sq. in.)

Answer 2

First off, it is good to hear you are okay! for you next question you have already considered the constant acceleration due to gravity, that is a good start!
The 'impression' your left in the ground doesn't really affect the impact, just the impacted, the opposite reaction as Newton put it. So lets look at the rest of the problem.
Our displacement was 30 ft, our approximate time .9 seconds. distance divided by our acceleration constant.
The total mass about 185 pounds, times the change in velocity divided by the change in time 33/.9 =37 time the mass is about 6783 pounds of force divided by the impact area 1356 pounds of instantaneous force per square ft of area.
If I remember that right, you should have been pretty tender for a while.

Answer 3

ok lets work this out so 185 pounds converts too about 83.92 kg.s so


If an object of mass of 83.92kg is dropped from a height
of 9.1 m, then the velocity just before impact is
9.899494936611665 m/s. The kinetic energy just before impact is equal to
= 245 J.

But this alone does not permit us to calculate the force of impact!

If in addition, we know that the distance traveled after impact is
d = 0.1 (im guessing because you didnt say , then the impact force may be calculated using the work-energy principle to be
Average impact force F = 2450N.

Answer 4

with the max. speed of thud

Answer 5

...yes.