Hi! I'm trying to solve a projectile motion question, so I want to know:
Since projectile motion is divided in X and Y components, does that mean I can swap "x" for "y" in equations when I'm solving for the Y part...
ie:
does Vfy^2 = Viy + 2ayy?
(V final in y direction squared = V initial in y direction plus parenthesis 2 times a in y direction times y end parentisis)
Wish we could input sub- and supersripts! :D
thanks :)
Since projectile motion is divided in X and Y components, does that mean I can swap "x" for "y" in equations when I'm solving for the Y part...
ie:
does Vfy^2 = Viy + 2ayy?
(V final in y direction squared = V initial in y direction plus parenthesis 2 times a in y direction times y end parentisis)
Wish we could input sub- and supersripts! :D
thanks :)
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Use ^ for superscript and _ for subscript. That's the standard notation.
First you forgot a square. V_f^2 = V_i^2 + 2a_x*x.
Normally in projectile problems, there is no acceleration in the x direction. a_x = 0. But in the case where you do have acceleration (due to wind for instance), then yes, you can write that equation separately for the x and y components.
And then Vf^2 = Vf_x^2 + Vf_y^2.
In the normal case (no acceleration in the x-direction, acceleration in the y-direction) it is true both for the total velocity:
Vf^2 = Vi^2 + 2a*d (d = change in vertical height)
and for the y-component of velocity
Vf_y^2 = Vi_y^2 + 2a*d
You get the 1st equation from the 2nd one by adding V_x^2 to both sides.
Edit: This equation is really conservation of energy in disguise. That's why the "x" is the vertical displacement. Work = force x displacement in direction of that force. So the work done by gravity is mg*h where h = vertical height. Final KE = initial KE + work done by gravity.
(1/2)mv_f^2 = (1/2)mv_i^2 + mgh
Cancel out the m's, multiply by 2 and you've got the v^2 equation.
First you forgot a square. V_f^2 = V_i^2 + 2a_x*x.
Normally in projectile problems, there is no acceleration in the x direction. a_x = 0. But in the case where you do have acceleration (due to wind for instance), then yes, you can write that equation separately for the x and y components.
And then Vf^2 = Vf_x^2 + Vf_y^2.
In the normal case (no acceleration in the x-direction, acceleration in the y-direction) it is true both for the total velocity:
Vf^2 = Vi^2 + 2a*d (d = change in vertical height)
and for the y-component of velocity
Vf_y^2 = Vi_y^2 + 2a*d
You get the 1st equation from the 2nd one by adding V_x^2 to both sides.
Edit: This equation is really conservation of energy in disguise. That's why the "x" is the vertical displacement. Work = force x displacement in direction of that force. So the work done by gravity is mg*h where h = vertical height. Final KE = initial KE + work done by gravity.
(1/2)mv_f^2 = (1/2)mv_i^2 + mgh
Cancel out the m's, multiply by 2 and you've got the v^2 equation.
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In projectile motion, gravity only acts downward, so gravity only deflects a projectile's vertical component (whether the vertical component is x or y or z depends on how you choose to orient your reference frame). If drag and wind are negligible, then the horizontal components of velocity are the same as their initial values. The reason projectile motion is parabolic is that the horizontal velocity is constant and the vertical velocity is constantly changing. (Although it gets a little cumbersome, I use ^ for superscripts and _ for subscripts. I think that's the convention.)
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