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plane
Oct5-05, 10:44 PM
i've noticed there are three formulas for displacement in projectile motion: dx=vixt, dx=vixt+0.5gt^2 and vfx^2=vix^2+2ad (same for in the y direction). im just confused as to when you use these three, can anyone please clear that up for me?

thanks!

hotvette
Oct5-05, 10:48 PM
Is there a specific problem you are trying to solve? You might want to check the equations again. You listed 2 different equations for dx.

plane
Oct5-05, 10:51 PM
Is there a specific problem you are trying to solve? You might want to check the equations again. You listed 2 different equations for dx.

these equasions relate to projectile motion, and i dont know which one of them to use for what kind of problem: yf=yi problems where they are kicked on level ground or those shot from a cliff. which one of these equasions do use when? and for what type of problems?

Gale
Oct5-05, 11:09 PM
OK, you know how you use kinematic equations normally? what you've written are various forms of the kinematic equations, all in the x direction. when you're doing projectile motion, you have axes, x and y. y is commonly vertical, x horizontal. when we fire a projectile, the only thing in common between the x and y planes are time. So, we do each separately.

Depending on what the conditions are, you use different equations. usually for a projectile, there is no acceleration in the x direction. so we use \delta x= v_{0x}t. when there's acceleration, we have to account for it. usually this is only in the y direction, because in the y direction there is acceleration due to gravity, g. so we use \delta y= y_{0} + v_{0y}t + \frac {1}{2} at^2 usually we use these in projectile motion problems. You can also use the last equation when you don't know time and you have a change in the velocities.

So, you have to understand what the equations mean and what you're given, and then you use what ever is most appropriate.

HallsofIvy
Oct6-05, 09:28 AM
The first, dx=vixt, is for constant velocity in the x-direction.

The second, dx=vixt+0.5gt^2, is for constant acceleration, g.
Be careful about applying this "also" to the y direction. Typically g is the acceleration due to gravity and it acts only in one direction!

The third, vfx^2=vix^2+2ad, is really a statement of "conservation of energy" since you can write it as 2ad= vfx^2- vix^2. If you multiply both sides by mass "m", you would have 2amd= vfx^2- vix^2 or ma d= (1/2)mvfx^2- (1/2)mvix^2. The left side is the work done by force= ma acting over distance d and the right hand side is the change in kinetic energy (assuming that vfx and vix are the final and initial velocities, respectively).

This does is NOT a vector equation- there is no "y" version. You must assume that vfx and vix are the speeds in whatever direction the object is moving.