Solving for θ Given Time in Projectile Motion

[misogynisticfeminist]

Hi, I'm trying to find out how a function relating $$\theta$$ and time together. this is because, in many cases of projectile motion, they do give us say, the angle at various spots of a particle's trajectory or velocities so that we solve for theta. Is it possible to create a function so that I know theta straight away just by substituting in the variable time??

 

[dextercioby]

What is the angle $\theta$...?Is it somehow connected to the double-slit experiment...?

Daniel.

 

[Nylex]

What do projectiles have to do with the double slit experiment?

 

[dextercioby]

They do,u see,it's all clear to me:projectiles,double slit experiments and Copenhagen vs.other QM interpretations...If u don't believe me read here and here


Daniel.

 

[SpaceTiger]

Hi, I'm trying to find out how a function relating $$\theta$$ and time together. this is because, in many cases of projectile motion, they do give us say, the angle at various spots of a particle's trajectory or velocities so that we solve for theta. Is it possible to create a function so that I know theta straight away just by substituting in the variable time??

And in case you're not doing the double slit experiment...

$$v_x=v_{x,0}$$
$$v_y=v_{y,0}-gt$$
$$x=x_0+v_{x,0}t$$
$$y=y_0+v_{y,0}t-\frac{1}{2}gt^2$$

Now I'm not sure what you mean by $$\theta$$ either, but it could be:

$$tan(\theta)=\frac{v_y}{v_x}=\frac{v_{y,0}-gt}{v_{x,0}}$$
$$tan(\theta)=\frac{y}{x}=\frac{y_0+v_{y,0}t-\frac{1}{2}gt^2}{x_0+v_{x,0}t}$$

If not, please clarify what you mean by $$\theta$$.

 

[dextercioby]

Aren't "x" and "y" perpendicular,so the angle between them [itex] \frac{\pi}{2} [/tex] and the tangent of this angle is $\infty$...?The same goes for velocity vector's projections onto the rectangular Oxy...

Daniel.

 

[SpaceTiger]

Aren't "x" and "y" perpendicular,so the angle between them [itex] \frac{\pi}{2} [/tex] and the tangent of this angle is $\infty$...?The same goes for velocity vector's projections onto the rectangular Oxy...

Uh, what? This isn't the angle between x and y. "x" and "y" represent the sides of a right triangle with one vertex at the origin. The angle is relative to the horizontal.

 

[dextercioby]

You're right.I think among your 2 options lies the OP's answer to his question,unless it's something which would defy logics.

Daniel.

 

[misogynisticfeminist]

And in case you're not doing the double slit experiment...

$$v_x=v_{x,0}$$
$$v_y=v_{y,0}-gt$$
$$x=x_0+v_{x,0}t$$
$$y=y_0+v_{y,0}t-\frac{1}{2}gt^2$$

Now I'm not sure what you mean by $$\theta$$ either, but it could be:

$$tan(\theta)=\frac{v_y}{v_x}=\frac{v_{y,0}-gt}{v_{x,0}}$$
$$tan(\theta)=\frac{y}{x}=\frac{y_0+v_{y,0}t-\frac{1}{2}gt^2}{x_0+v_{x,0}t}$$

If not, please clarify what you mean by $$\theta$$.

hey, that helped, thanks alot.

I was not looking for the double-slit exp. though lol. I wouldn't have posted that in the classical physics forum. and that helped too, btw.

: )