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jagguy2

falling bodies physics

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hi, what is the formula to calculate the velocity of a falling body from a height ? I want to X value and Y value at any time t. i have 2 formulas 1)after falling from a height h, the vertical velocity is sqrt(2*g*h) 2)Vt = square root of (2mg/ÁACd) Vt = terminal velocity, m = mass of the falling object, g = gravitational acceleration, Cd = drag coefficient, Á = density of the fluid the object is falling through, and A = projected area of the object. for example q1)How do i work out the vertical/horizontal velocity of an person walking 5km off a cliff that drops 100m. q2)if i drop a feather and a rock from 10m, surely the rock will get there quicker.

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Quote:
Original post by jagguy2
q2)if i drop a feather and a rock from 10m, surely the rock will get there quicker.
They should have covered this in a basic physics class, but the short answer is that they fall at the same speed (neglecting air resistance) - see Galileo's experiment. In practice though, the feather has such high drag as opposed to its weight that air resistance will cause it to flutter slowly.

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ok i forgot the basic physcis class as it was a few years ago(but thanks for thelink).Anyway I was really after an example for

for example
q1)How do i work out the vertical/horizontal velocity of an person walking 5km off a cliff that drops 100m.

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Quote:
Original post by jagguy2
ok i forgot the basic physcis class as it was a few years ago(but thanks for thelink).Anyway I was really after an example for

for example
q1)How do i work out the vertical/horizontal velocity of an person walking 5km off a cliff that drops 100m.

Depends on what factors you want to come into play (wind resistance? weight distribution?). You mention terminal velocity, do you know what that is and how it comes into play? Also, idealised horizontal components of velocity remain constant, do you want a horizontal component capable of change in direction/magnitude?

q1 is nebulous...
How does one walk 5 km off of a cliff? That's like someone walking 3 miles in air before falling 6% of 1 mile.

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Original post by jagguy2
the person is walking 5km. they come to a cliff and keep going and fall of course.

it is for a computer game.

Ok, he walked 5km before coming to the edge of the cliff. So? What was his velocity before the fall? Are we to just assume straight free fall? I could walk a infinte amount of distances and fall off a cliff an infinite amount of ways.

edit: Do you mean 5 km/hour ?

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hi,

what i wanted was the mathematics and not just an
answer to be honest.

q1)TO make it more interesting say they are moving 50km/hour off a cliff of 100M .

I want an example of calculating the velocity , horizontal velocity and vertical velocity components every t secs.
I just want to see the mathematics involved as well as the answer of course.

I am modeling the fall on a computer and not interested in all the variables like weather, air resistance, wind speed on the day, etc just the basic mathematics.

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Um... no...

If you really need it in an equation, it would be Y = gt^2, assuming zero vertical velocity at t=0, and g=-9.8m/s/s (generally the rate of gravitational acceleration). Air resistance is VERY rarely calculated in a game setting, and often only takes the form of a velocity cap. However, in a game setting, you almost always use velocity vectors, in which case you apply downward acceleration by adding a down vector of magnitude 9.8m/s to the objects velocity vector each second (or every tick). You then get your position by adding the velocity vector to the position vector to get the new position for each tick/frame.

Additionally, horizontal velocity does not affect vertical velocity. If his horizontal velocity when he walked off the cliff was 5km/h, then his horizontal velocity remains 5km/h for the fall (unless you are calculating air resistance, which I highly recommend against).

So, to get the velocity, it would be:
vY = g*t
vX = 5km/h

For position, it is:
Y = g*t^2
X = 5*t

However, vectors work better for simulations.

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