Posted on 07/14/2009 9:33:55 AM PDT by rdl6989
A man died after his car plunged 600 feet off the edge of the Grand Canyon's South Rim, authorities said Tuesday. The Arizona park's regional communications center received several reports of a car driving off the edge about 6 a.m. Monday, according to a written statement.
"Upon arriving at the scene, investigators found tire tracks leading to the edge behind the Thunderbird Lodge and received reports of a single occupant in a blue passenger car driving over the edge," the statement said.
Rescue personnel descended on ropes and found the vehicle about 600 feet into the canyon. The man's body was recovered shortly afterward, the statement said.
The incident occurred near the El Tovar hotel in a village on the canyon's South Rim, park spokeswoman Shannan Marcak said.
Authorities have not ruled the death a suicide, she said. "It has not been ruled anything at this time."
The statement said the National Park Service is investigating. Typically, Marcak said, such investigations take at least a few days.
(Excerpt) Read more at cnn.com ...
Can you run that through the graph program and show the parabolic track, as soon as the void opened, downward motion begins and as you said the horizontal impact distance would be roughly 2/3 of the edge of the clifftop; a sheer drop would put him about 45 feet past the cliffwall in the first second — arcing out to the terminal point.
It could grip it by the husk.
“Not exactly, forward speed will affect the trajectory, but not how quickly he hits the bottom of the canyon. So if his forward speed is sufficient, and the other side low enough he could make it to the other side.”
I get questions like this when I explain bullet drop to my friends. Questions like, “Does it make a difference if I use a 68 grain bullet instead of 55 grain?”
Speed is irrelevant, 32 feet/sec, gravity is a wonderful thing...
At 1500 fps you cannot ignore drag. If you shoot straight down, air resistance initially will certainly be a lot greater than gravity. When drag equals gravity, you are at "terminal velocity".
To a first order, for a car traveling at 40 MPH drag is a lot less than gravity.
I took physics 1.01 with Professor Resnick and he had a famous lecture demonstation, with a rifle pointed at a stuffed toy monkey suspended about the stage by an electro-magnet. He would throw a switch and the current to the magnet would be cut at the same instant that the rifle was discharged. The monkey would be hit after falling about three feet, even though the rifle was initially aimed directly at him. The monkey and bullet underwent the same vertical acceleration.
The Husk...?.....uuuhhhhhhh....wouldn’t it be too big to lift up?
Oh! You owe me a new keyboard!!!! :-)
Gimme a big ole guardrail and I'm OK....but they don't do that sort of thing there, I guess.
As for calculating his, er, terminal velocity, nobody seems to be taking into account that he could have had his foot on the brake the entire way down. I would have.
If our government cared about us, they would make the car companies put parachutes on the cars.
If he was going 120 mph wouldn't it take longer than if he was going 20? what about an arc?
Regardless, it would be between 10 and 16 seconds. Damn that is a very long time to think about your impending doom.
Terrible. The 40 mph governs how far out in the canyon he will travel over the time taken to fall 600 feet.
The time required to drop 600 ft. is driven by gravity, coupled with the initial downward speed of the car.
From your freshman physics class, you will recall that
d = d0 + v0t + 1/2 agrav t2
Since he drove off the edge horizontally, his initial downward speed v0 is approximately zero, as is his initial distance, d0.
Thus,
d = 1/2 agrav t2
d = 600 ft, and agrav = 32.174 ft/sec2.
So we can solve for t to get a "hang time" of about 6.1 seconds.
Over that time he would travel approximately 360 feet out from the edge of the canyon.
lmao
His speed doesn't matter, because it was horizontal. What counts is the vertical acceleration of 32 feet per second per second. This works out to a bit over 6 seconds, ignoring air friction. Call it 7, and you'll probably be pretty close.
Of course, this assumes he flew to 600 feet down, and stopped abruptly. He probably bounced a few times, which would have taken longer.
The horizontal and vertical components of the motion of the car are independent. It will fall at the same rate whether it has 0 forward motion or 120 mph forward motion.
I can’t stand Susan Sarandon and the world would be better off if she did jump that car, but I always liked Geena Davis for some reason.
If he dropped 600', it would take the same number of seconds at 40mph off the cliff as it would at 65mph or 105mph off the cliff.
The acceleration [downward] of gravity is a constant 32ft/sec/sec.
Assuming he drove off the cliff exactly horizontally the equation is D=1/2 at2, where D = 600', and a=32'/sec/sec. Solving for t [time] you get 6.12 seconds - far to long to contemplate your mistake.
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