a projectile near the earth's surface moves in a parabola. so what is the equation for it's path? it opens down, so it is [imath]y=-ax^2+bx+c[/imath], but what are a b and c? if we put one zero at x=0, c goes away. that leaves [imath]y=-ax^2+bx[/imath]. so, what are a and b?

Give me the positions of two more points of your parabole and we can give you a and b

 

if you understand what is show here you will know:

http://www.glenbrook.k12.il.us/gbssci/phys/chemphys/reviews/u7review/u7ans3.html

the position of a projectile divided into x and y components is:

 

[Math] x=v_{0x} t [/math]

[math] y=v_{0y} t - \frac{1}{2} g t^2 [/math]

 

solve for [math]t[/math] and plug in:

 

[math]y= \frac{v_{0y}}{v_{0x}}x - \frac {1}{2} g (\frac{x}{v_{0x}})^2 [/math]

 

then, you know that

[math]\tan \theta_0 = \frac {v_{0y}}{v_{0x}}[/math]

and the x component of the initial velocity is

[math] v_{0x} = v_0 \cos \theta_0 [/math]

plugging those in, you get

 

[math] y= (\tan \theta_0)x - \frac {g}{2(v_0 \cos \theta_0)^2} x^2[/math]

 

you can see [math]a[/math] and [math]b[/math] pretty clearly like that i think

 

hopefully this helps... first time i used latex... maybe it'll get less annoying when i get used to it

is that the actual anser? it makes sense, but if you just made it up on the spot i want confirmation.

 

nice [math]\LaTeX[/math] for a firsttimer

what do you mean? i've definitely seen it done this way before, and i'm not making it up off the top of my head or anything. i'm almost completely certain that it works.

look here:trajectory wikipedia

if you scroll down 4/5 of the way or so, look for equation 10- it's exactly the same thing with different variables.

ok. just making sure.

 

i didn't want to use it on a test and it be total crap.

 

ty