How do I derive the projectile parabola equation using equations of motion?

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The discussion focuses on deriving the projectile parabola equation using the six equations of motion for constant acceleration. Key equations mentioned include v2 = v1 + at, x2 = x1 + 1/2(v1 + v2)t, x2 = x1 + v1t + 1/2at², and v2² = v1² + 2a(x2 - x1). The remaining two equations are not specified, but the derivation utilizes acceleration due to gravity (a = -g) in the vertical direction and zero acceleration (a = 0) in the horizontal direction to establish the parabolic path.

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nameVoid
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write out the six equations of motion in 1 dimension for constant acceleration and from those derive the equation y(x) for the parabolic path of a projectile.


I have
v2=v1+at
x2=x1+1/2 (v1+v2)t
x2=x1+v1t+1/2 at^2
v2^2=v1^2+2a(x2-x1)

not sure of the remaining 2 and the derivation
 
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hi nameVoid! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)
nameVoid said:
I have
v2=v1+at
x2=x1+1/2 (v1+v2)t
x2=x1+v1t+1/2 at^2
v2^2=v1^2+2a(x2-x1)

not sure of the remaining 2 and the derivation

hmm … i have no idea what the other 2 https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations are supposed to be :confused:

but anyway you should be able to get the projectile parabola by using those four …

with a = -g in the y direction, and a = 0 in the x direction :smile:
 
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