Math people...

[xer0s]

Ok, lets say I have this thing that shoots stuff. Now, lets say this thing shoots an object 970 yards, and is shot at a 60 degree angle. Is it possible to figure how high the object went when it was shot? And if not, what other factors would need to be known to figure this out?

P.S. Props goes out to DooMer for hooking my account back up.

[phantasmagoria]

i think you need to know the force behind the object when it's fired.

[Arkleseizure]

I think your variable is how much force was behind the object. Also, the mass deals with the equation.

[stocktroll]

uh, dont you mean "physics" people?

[losCHUNK]

and how high you was when you was jacking off ?

[tnf]

Ok, lets say I have this thing that shoots stuff. Now, lets say this thing shoots an object 970 yards, and is shot at a 60 degree angle. Is it possible to figure how high the object went when it was shot? And if not, what other factors would need to be known to figure this out?

P.S. Props goes out to DooMer for hooking my account back up.

Ignoring air resistance?

[Freakaloin]

Ok, lets say I have this thing that shoots stuff. Now, lets say this thing shoots an object 970 yards, and is shot at a 60 degree angle. Is it possible to figure how high the object went when it was shot? And if not, what other factors would need to be known to figure this out?

P.S. Props goes out to DooMer for hooking my account back up.

the answer is 4...

[phantasmagoria]

Ok, lets say I have this thing that shoots stuff. Now, lets say this thing shoots an object 970 yards, and is shot at a 60 degree angle. Is it possible to figure how high the object went when it was shot? And if not, what other factors would need to be known to figure this out?

P.S. Props goes out to DooMer for hooking my account back up.

Ignoring air resistance?

wouldn't air resistance be a universal constant and therefore cancel itself out in the vertical and horizontal planes? i haven't done this stuff for years so i'm probably wrong..

Thinking about it you'd definatly need the mass, also the initial acceleration of the object...possibly

[[xeno]Julios]

since you know the range of the trajectory as well as the angle, you should be able to calculate the max height, since trajectories within an environment of constant gravity have typical shapes.

In other words, there is only one trajectory shape that fits a given angle and range.

The key is to quantify a specific feature of this shape: the height of the "rainbow".

[Fender]

You don't need to know mass or "force" or acceleration. 60 degrees and 970 yards is all you need to know.
start with x = 1/2 * a * t ^ 2

[phantasmagoria]

is it actually possible to do it that way or are you theorising julious?

[Fender]

a = 9.8 m/s^2
a = 10.72 y/s^2
Now multiply by sin(60) or cos(60), i forget which and I've had a couple too many beers to bother figuring it out. It is easy, though.

[[xeno]Julios]

is it actually possible to do it that way or are you theorising julious?

theorizing - but i'm pretty confident i'm right

[phantasmagoria]

fuck of course, i hated equations of constant acceleration; infact i hated all mechanical maths

[[xeno]Julios]

since you don't have time, mass, or force, you're gonna hafta set up some simultaneous equations which cancel out those factors, leaving only angle and range.

Range = distance traveled along x axis, which is total time multiplied by x-velocity

x-velocity is Velocity multiplied by cos theta, so in this case, x-velocity = 1/2 V (since cos 60 = 1/2)

y-velocity is 0 halfway through the trajectory, so you can fumble around with acceleration equations.

There's probably a handy formula that does all this - might be very tricky to derive from first principles.

[Fender]

again, why do euros say "mathS" vs "math"???

[phantasmagoria]

because it's the proper way to say it

[Fender]

bah

[Arkleseizure]

Go shit on yourself phantasmagoria.

[[xeno]Julios]

K i think i mighta got it:

based on this page:

http://hyperphysics.phy-astr.gsu.edu/hb ... 0&h=0#tra5

Range = v times sqrt(2h/g)

and

Height = (v*v*sin theta*sin theta) /2g

i rearranged the equations, plugging in what I knew (range and theta), solved for v squared, and used a simultaneous equation to get this:

4*h*h*g/0.75 = 960*960g



solving for h i get:

415.69 yards

Not sure if this is correct though...

[[xeno]Julios]

k i checked the answer by calculating range, given the max height, and it worked out.

[[xeno]Julios]

the more general equation:

max height = sqrt (sin theta * sin theta * range * range / 4)

[DRuM]

again, why do euros say "mathS" vs "math"???

maths short for mathematics. Plural because it's to do with numberS. Also because we invented proper english.

[Guest]

Don't forget to convert it to fucking meters, for fuck's sakes.

[Pauly]

Americans always like to drop letters. Color instead of Colour, Armor instead of Armour.

They are so fucking lazy they can't even say full words. No wonder they're fat bastards, sorry bastads.