# Jumping Into Fire Net

Matt
A person jumps from a fourth story window 15 m above a safety net. THe jumper stretches the net 1.0 m before coming to rest. What was the deceleration experienced by the jumper?

Equation:

x = x0 + v0t +.5at^2

15 m = 1 m + 8.6 m/s(2s) + .5(a)(1.75s^2)

15m = 1 m + 17.2 m/s^2 + .875s^2(a)

-3.2 m = .875 s^2

-3.66 m/s^2 = Deceleration

Does this look right? I'm not sure about the 1 m as the final position...

Matt
Wait, I already see a problem, forgot to square the time...

Equation:

x = x0 + v0t +.5at^2

15 m = 1 m + 8.6 m/s(2s) + .5(a)(1.75s^2)

15m = 1 m + 17.2 m/s^2 + 1.53s^2(a)

-3.2 m = 1.53 s^2

2.09 m/s^2 = Deceleration

Does this look right? I'm not sure about the 1 m as the final position...That seems like an awfully slow deceleration for jumping out of a window.

enigma
Staff Emeritus
Gold Member
Originally posted by Matt
A person jumps from a fourth story window 15 m above a safety net. THe jumper stretches the net 1.0 m before coming to rest. What was the deceleration experienced by the jumper?

Equation:

x = x0 + v0t +.5at^2

15 m = 1 m + 8.6 m/s(2s) + .5(a)(1.75s^2)

15m = 1 m + 17.2 m/s^2 + .875s^2(a)

-3.2 m = .875 s^2

-3.66 m/s^2 = Deceleration

Does this look right? I'm not sure about the 1 m as the final position...

You've got a few errors, and I'm not sure where you got all of your values.

Where did the 8.6m/s , the 2s, and the 1.75s^2 come from?

You're also making mistakes with the units.

'8.6 m/s * s' does not have units of 'm/s^2', it is 'm'

Your tipoff that you are making mistakes comes in the line where you have -3.2m = .875 s^2

those units don't equal each other, therefore you know you must have made a mistake somewhere.

To do this problem correctly, you need to break it into 2 parts.

First part is the freefall. You know starting and ending positions (15m and 0m above the net), you know starting velocity (0m/s), and you know acceleration (-9.8m/s^2). You use those numbers to solve for the velocity before the guy hits the net.

The second part is the deceleration. You have starting and ending positions (0m and -1m), you know starting velocity (from part I), and you know final velocity (0m/s... he is stopped at the bottom). You use those values to solve for acceleration.

Matt
Revised: (I always have trouble seeing problems in two distinct parts)

v^2 = v^20 + 2a(x-x0)
v^2 = 0 m/s + 2(9.80 m/s^s)(15m -0m)
v^2 = 294 m^2/s^s
Squareroot(v^2) = Squareroot(294 m^2/s^2)
v = 17.14 m/s

v^2 = v^20 + 2a(x-x0)
0 m^2/s^2 = 17.14 m/s + 2a(-1 - 0)
0 m^2/s^2 = 294 m^2/s^s -2a
-294 m^2/s^2 = -2a
a = 147 m/s^2

Thats more than 14 gs!!! Can that be right? Wouldnt the person be dead?

enigma
Staff Emeritus
Gold Member
15 g's.

The net is slowing him down to zero velocity in 1/15th the distance.

Not a very good net - He would probably be hurt, but not neccessarily killed.

15 g's is enough to cause damage, but it won't last for a very long period of time. Aircraft ejector seats pull over 100 g's for a split second.

Is the result reasonable? If you don't have a clear mental picture of meters, convert to feet. This guy is falling from a fourth (probably closer to 5th) story window, and stopping in 3 feet. That's a pretty stiff net.