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Subject:
Foot per second spped of Bungi Jumping
Category: Science > Physics Asked by: bonita111-ga List Price: $2.00 |
Posted:
13 Jul 2006 12:09 PDT
Expires: 12 Aug 2006 12:09 PDT Question ID: 746026 |
What speed will a person reach, who free falls 111 meters? (ex. Bungi Jump at Victoria Falls) |
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There is no answer at this time. |
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Subject:
Re: Foot per second spped of Bungi Jumping
From: myoarin-ga on 13 Jul 2006 14:35 PDT |
I doubt that this can be answered precisely, since I doubt that there is just one standard for bungi elastics or when they will start to break the speed of fall. Also, we need a better definition of "free fall 111 meters": 111 mtrs prior to the elastics starting to reduce speed? (That would be easy.) 111 mtrs maximum drop - the point when the elastics stop descent. |
Subject:
Re: Foot per second spped of Bungi Jumping
From: oorca-ga on 13 Jul 2006 14:58 PDT |
It is contingent on both the weight and attitude ( ie curled as a ball or limbs extended) of the the jumper, as well as the definition of 111 meters. |
Subject:
Re: Foot per second spped of Bungi Jumping
From: steph53-ga on 13 Jul 2006 20:06 PDT |
I'd freak out!!! Steph53 |
Subject:
Re: Foot per second spped of Bungi Jumping
From: toufaroo-ga on 14 Jul 2006 11:52 PDT |
The other guys are all unfortunately correct, because drag will play a significant part in your final speed...But here's an idea of how you can calculate the max speed, assuming no aerodynamic drag, no bungee elasticity, etc. Using one of the equations of motion for constant acceleration,... deltay = vo*t + 1/2 a*t^2, and plugging in deltay = 111 m, vo = 0, a = 9.81 m/s^2, we get t = 4.75 seconds. So, you'd be in freefall for 4.75 seconds. Next, using another equation of motion v = vo + a*t, and plugging in vo = 0, a = 9.81, t = 4.75, we get v = 46.7 m/s So, the FASTEST you could possibly be travelling from a fall of 111 m here on earth is 46.7 m/s or approximately 104 miles per hour. You will be going considerably slower because drag will slow you down somewhat. Of course, at the exact moment that the bungee reaches its maximum stretch and is about to recoil, you will be travelling 0 mph. |
Subject:
Re: Foot per second spped of Bungi Jumping
From: toufaroo-ga on 14 Jul 2006 11:54 PDT |
Sorry - forgot to include that 46.7 m/s = 153.2 ft/s |
Subject:
Re: Foot per second spped of Bungi Jumping
From: knowitawe-ga on 21 Jul 2006 20:44 PDT |
If you free fall 111m, you will reach a maximum speed of 46.6 m/s at 111m. However, you aren't free falling if you bungee jump and survive ;). You free fall to a maximum speed when you reach the end of the bungee and then begin decreasing speed until you reach the end of the fully stretched bungee where you reach zero 0 m/s and reverse direction. So, if the bungee causes you to stop and reverse direction at 111m, then maximum speed was actually reached at the length of the unstretched bungee (i.e., the length of the bungee when hung without your weight attached) So, pick your bungee length and max speed: 20m 19.8m/s max 40m 28 m/s max 60m 34.3 m/s max 80m 39.6 m/s max 100 44.3 m/s max One web site indicates that the bungee at victoria falls stretches 3 times it's length. So, if you go the full distance of 111m and that's three times the length of the bungee, the bungee length is 37m. According to the eThat means the maximum speed is 26.9 m/s. |
Subject:
Re: Foot per second spped of Bungi Jumping
From: knowitawe-ga on 21 Jul 2006 20:56 PDT |
To complete the previous comment and give credit for it's calc: Using the calculator at http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html , the maximum speed reached by a Victoria Falls jumper going a distance of 111m with a 37m bungee cord would be 26.9m/s. Note: If the 37m bungee breaks, the "terminating velocity" ;) will be something between 26.9m/s and 46.6m/s. Happy jumping.... |
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