Falling Bodies ```Name: Mike Status: other Age: 30s Location: N/A Country: N/A Date: 2000-2001 ``` Question: I'm a Police Officer in Ada, Oklahoma. Last week we had a suicide, where the Individual jumped to his death. He weighed approximately 160 pounds and fell 51 feet to his death. During the investigation, it was pondered on how long he was in the air before he landed. It appears that he jumped out from where he was standing 3-5 feet. Can you please help? Doesn't an object reach terminal velocity within 32 feet of falling? One Officer believes it took approx 1.5 seconds, and myself and the medical examiner where thinking it took approx 2 to 2.5 seconds. thank you Mike Baker Replies: Terminal velocity makes the problem a lot more tricky, but fortunately, this doesn't look like terminal velocity had much of an effect in this case. For a typical person, terminal speed is 60 m/s or approximately 135 miles per hour. A person has to fall over 400 yards before you really need to start taking this into account. The 32 feet you remember is the acceleration to gravity. A falling object increases its velocity by 32 feet per second per second it falls. Also, the fact that he jumped out doesn't really affect the equations. If he jumped up or down that would change things slightly, but probably not enough to change the answer by more than a very small fraction of a second. First we convert feet to meters: 51 feet = 51f/3.28f/m = 15.55 m. So, from physics we can use the formula for constant acceleration: x - x0 = v0 * t + 1/2 a*t*t 51/3.28 = 0 + 1/2 * 9.8 * t * t 15.55 = 4.75 * t * t 3.273 = t*t t = sqrt(3.273) Time it took then is: 1.81 seconds. Thanks, Eric Tolman By my calculation, it would have taken about 1.8 seconds. t = sqrt(2x/a), where 'x' is the distance in meters and 'a' is 9.8 m/s/s, the acceleration of gravity. When the guy landed, he would have been moving at around 17 meters/second or 38 mph. I doubt this is close to terminal velocity for a body. I've heard that skydivers routinely get over 100 mph. If so, then it's ok to ignore air resistance in this calculation. Tim Mooney If one simply assumes that the object fell 51 feet, without air resistance and zero initial velocity, at least in the vertical direction, then the object would have been in the air for 1.78 seconds. Distance= Acceleration due to gravity x time x time/2 51=32x t squared/2 or the time= 1.78 seconds. The other factor such as initial velocity in the vertical direction would reduce the time in the air. Achieving terminal velocity, would also increase the time in the air. Unless the poor chap fell with his body parallel to the ground, I doubt that terminal velocity played a role in this scenario. Dr. Myron First of all, science aside, my condolences to the victim and his family. Someone takes his own life only when something, somewhere, has gone terribly wrong. Back to science. You can probably ignore the effects of wind resistance for a human body falling 51 feet. To reach terminal velocity, the wind drag force must equal the weight of the falling body. You have to be going pretty fast for this to happen. Without wind resistance, the speed of a falling object increases by 32 feet per second each second. (This is probably where you heard the "32 feet" number.) It takes a little calculus to figure out how fast an object will be going after falling for 51 feet, and how long it takes to fall that distance. Assuming that it starts from a dead stop, the formulas are: velocity = v = sqrt(2gh) time = t = sqrt(2h/g) where h is the height (distance the object falls) and g is the gravitational acceleration, (32 feet/sec)/sec. From these formulas, an object would take 1.8 seconds to fall 51 feet, and would be traveling 57 ft/s (40 mi/h) when it gets there. That doesn't sound very fast, does it? It should put into perspective how deadly car crashes can be without all those modern safety features. Richard E. Barrans Jr., Ph.D. Assistant Director PG Research Foundation, Darien, Illinois Click here to return to the Physics Archives

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