Projectile motion problem

[FaraDazed]

Homework Statement

Points P and Q are in the same horizontal plane at a distance d metres apart. A projectile is shot from P in the vertical plane through PQ towards Q with a speed v m/s at an angle θ. Simultaneously another particle is shot from Q in the same vertical plane towards P with a speed of 2m/s at an angle of elevation α .

Find an expression for the distance between the two particles when one is vertically above the other one and hence prove that they collide if $sin \alpha = \frac{1}{2} sin\theta.$

If $\theta= \frac{1}{3}\pi$ prove that they collide at a point above the level of PQ if...
$$v^2 > \frac{gd(\sqrt{13}-1)}{6\sqrt{3}}$$

Homework Equations

equations of motion

The Attempt at a Solution

I just don't know how to approach this one at all. I don't want to just be given the answer, just some advice or help in how to approach the problem and any assumptions would be very very much appreciated.

 

[Mentz114]

I cannot understand the setup. Is PQ the line joining P and Q ? Also, if P and Q are in a horizontal plane, how can P or Q be vertically above the other ?

 

[FaraDazed]

I cannot understand the setup. Is PQ the line joining P and Q ? Also, if P and Q are in a horizontal plane, how can P or Q be vertically above the other ?

I think that's what it means when it says PQ - being the line/distance joining the two. And, sorry about the second bit that was my fault copying the question down; P and Q are points on the plane, its the particles that are fired from P and Q that are/will be above each other.

Thanks

 

[Mentz114]

OK. You can find the the time when the projectiles are above each other by solving x_P=x_Q where

x_P= vt cos(θ) and x_Q= d-vt cos(α)

Does this help ?

[edit] I made a mistake, I've corrected it.

 

[FaraDazed]

Thanks, isn't the x cordinate of a projectile in this situation generally speed*t as in $v\cos(\theta) t$ though?

If i set them both to equal each other though I would get stuck as there is nothing I can do to manipulate it, is there?

$$v\cos(\theta)=d-v\cos(\alpha)$$

 

[Mentz114]

I'm sorry you got caught by my mistake. I've edited my post so you can find the time now, and plug it into the equations for y to get the heights.

You get the time from $vt\cos(\theta)=d-vt\cos(\alpha)$

 

[FaraDazed]

OK thank you.

I am stuck with the algebra side of it then I think as I cannot see how to extract t from that. Plus not knowing either of the angles is confusing the hell out of me, I suppose if t is found then the alpha angle could be found as it states the speed for that one (2m/s).

 

[Mentz114]

The value of t when the projectiles are at the same x is

$vt\cos(\theta)=d-2t\cos(\alpha)\ \ \rightarrow\ vt\cos(\theta)+2t\cos(\alpha)=d\ \ \rightarrow t_s=\frac{d}{v\cos(\theta)+2\cos(\alpha)}$

The y values are

$y_P=vt\sin(\theta),\ y_Q=2t\sin(\alpha)$ put $t_s$ into these and then subtract them.
corrected in my next post

You need to do the algebra.

 

[FaraDazed]

The value of t when the projectiles are at the same x is

$vt\cos(\theta)=d-2t\cos(\alpha)\ \ \rightarrow\ vt\cos(\theta)+2t\cos(\alpha)=d\ \ \rightarrow t_s=\frac{d}{\cos(v\theta)+2\cos(\alpha)}$

The y values are

$y_P=vt\sin(\theta),\ y_Q=2t\sin(\alpha)$ put $t_s$ into these and then subtract them.

You need to do the algebra.

Thanks for your help. :)

With the y values, doesn't gravity need to be accounted for?

 

[Mentz114]

Thanks for your help. :)

With the y values, doesn't gravity need to be accounted for?

Yes. Sorry I'm making a lot of mistakes. The y equations (assuming that the initial height is 0) are

$y_P=\int(v\sin(\theta)-gt) dt=vt\sin(\theta)-(1/2)gt^2$ and $y_Q=\int(2\sin(\alpha)-gt)dt=2t\sin(\alpha)-(1/2)gt^2$

So the difference in height is $vt\sin(\theta)-2t\sin(\alpha)$ and the Y-values are equal (for t>0) only if $v\sin\left( \theta\right) \,=2\,\sin\left( \alpha\right)$ which is *not* what is given for this condition.