Dark Bit Factory & Gravity
PROGRAMMING => General coding questions => Topic started by: Tetra on March 11, 2007
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I know this may seem like silly question, but I can seem to get it right and wondered if anyone could make it cleare to me.
I know that on earth that gravity has an effect on an object close to the surface at about 9.807m/s2
In terms of actually applying gravity on something what does that mean, that every second that passes the falling object has gained the speed of 9.807 every second?
It seems pretty insane if an object is travelling at 220miles an hour (98.07m/s) just after 10s of free fall if there was no air resistance.
That can only lead me to beleive I just dont get it :D I've forgotten all that I did at school, so if anyone would be able to remind me of the correct way to do it that would be great :)
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That's about right, but the resistance from air increases exponentially with velocity which means that with air resistance taken into account the faster something is falling then it's acceleration drops more and more until the force of the air resistance is equal to the force of gravity and you have terminal velocity so it might never even reach 220mph (depending on the mass and surface area of the object).
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Here are some equations that may help:
http://www.mansfieldct.org/schools/mms/staff/hand/Lawsgrav.htm
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:cheers: guys thanx for your replies, very helpfull :)
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Typical skydivers reach terminal velocity of only 125mph = 200kph, 55m/s, around 5.5s of acceleration.
v=u + at
s=ut + 0.5at^2
v^2 = u^2 + 2as
s = t(u+v)/2
Jim