Damnit..does anyone know the projectile equation (physics)?

[Nullity]

You give it an initial velocity and an angle and it'll give you the distance it has traveled.

Oh..the projectile also has a parabolic shape and must start from the ground. (like a cannon)

I need it!!
Thanks,
Null

 

[FrontlineWarrior]

split into x & y vectors my friend.

 

[Nullity]

Nevermind..found it:

projectile range = (v^2 * sin (2*theta)) / g

 

[Weyoun]

i wonder how they actually figured out that equation... im the type of person who can't remember anything they cannot reproduce, so im screwed then

 

[thEnEuRoMancER]

Weyoun:

Split velocity v in two components

vx=v*cos(theta)
vy=v*sin(theta)

Time it takes for the projectile to reach the highest point

t=vy/g

Time it takes to reach the highest point and fall back down is two times t. In this time the projectile travels the following horizontal distance:

d=vx*2*t=v*cos(theta)*2*v*sin(theta)/g = v^2*2*sin(theta)*cos(theta)/g

2*cos(theta)*sin(theta)=sin(2*theta)

so

d=v^2*sin(2*theta)/g

Deriving equations is a good way of practicing physics.

 

[rmeijer]

If you want a more robust derivation, I'd suggest using a "free body diagram". Look at the object in terms of what forces are acting on it (i.e. F = ma). Do this exercise in both the horizontal and vertical planes.

Since a is simply dv/dt, derive your equations w/r to time..... that will give you an equation with v = f(t).

Make sense?

 

[Mday]

splitting it into components is the way to go, for SOME problems.

unless you want the equation to solve things... equations, blah...