Can a Skydiver Survive a Free Fall Without a Parachute?

[Xman1120]

Homework Statement

For a brief moment, the human body can withstand accelerations of up to 75g. If a sky diver is unlucky enough to have a parachute fail, and hits the ground with a terminal velocity of 50 m/s, what is the minimum distance over which he can come to a rest (assuming constant acceleration) and survive. This occasionally happens if one is fortunate enough to land in thick brush, deep snow and/or on a steep hill.

Homework Equations

[tex]\Delta[/tex]x = Vi + 1/2 * a *t^2
Vf^2-Vi^2 = 2*a* [tex]\Delta[/tex] x

The Attempt at a Solution

Without having a height or a time, my assumption would be to use Vf^2-Vi^2 = 2*a* [tex]\Delta[/tex] x , but when doing this I assume the acceleration would be 9.8 m/s^2 , the terminal velocity being Vf which is 50 m/s, however when i calculate and solve for x i get 127.55 m, yet I don't believe that's the minimum. where am I going wrong?

 

[hotvette]

You have the initial velocity (50 m/s), final velocity (0 m/s), and acceleration (75g). From this information you should be able to determine distance.

 

[Xman1120]

Then I'm assuming that after my calculations, that would be the minimum distance?

 

[hotvette]

Yes. If the max acceleration the human body can stand is 75g, the distance associated with a body slowing down to rest would be the minimum distance. Greater distances would imply lower acceleration.

 

[rwisz]

Your approach is correct, but your assumption for the acceleration is incorrect. The acceleration experienced by the skydiver will not be 9.8 m/s^2, as they are not in free fall with a parachute. Instead, they are experiencing a much higher acceleration of 75g, which is equivalent to 735 m/s^2. Therefore, when using the equation Vf^2-Vi^2 = 2*a* \Delta x, you should use 735 m/s^2 for the acceleration.

Additionally, the initial velocity (Vi) in this case would be 0 m/s, as the skydiver is starting from rest. Therefore, the equation becomes Vf^2 = 2*a* \Delta x. Plugging in the given values, we get:

50^2 = 2*735* \Delta x
\Delta x = 0.034 m

This means that the minimum distance over which the skydiver can come to a rest and survive is only 0.034 meters, which is a very short distance. This is due to the extremely high acceleration experienced by the skydiver. So even if they were to land in thick brush, deep snow, or on a steep hill, they would still need to come to a stop over a very short distance in order to survive.