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Jumping Into Fire Net

  • Thread starter Matt
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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

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Science Advisor
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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

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Science Advisor
Gold Member
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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.
 

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