1. ## Relative Velocity

3 particles A,B,C are projected from a point above the ground level.A launches horizontally with a velocity of U and B, was launches $90-\theta$from vertical with the velocity of W and C was launches vertically with the velocity of V.Using the relative velocity, If $\frac{cos\theta}{U}+\frac{sin\theta}{V}=\frac{1}{W }$,Show that The particles A,B,C are in a straight line??

In this case Give me hint to use Relative velocity>>><<<<

2. ## Re: Relative Velocity

Originally Posted by srirahulan
3 particles A,B,C are projected from a point above the ground level.A launches horizontally with a velocity of U and B, was launches $90-\theta$from vertical with the velocity of W and C was launches vertically with the velocity of V.Using the relative velocity, If $\frac{cos\theta}{U}+\frac{sin\theta}{V}=\frac{1}{W }$,Show that The particles A,B,C are in a straight line??

In this case Give me hint to use Relative velocity>>><<<<
looking at the velocity vectors

\begin{align*}&v_A=\{U,0\} \\ &v_B=\{W \cos(\theta),W \sin(\theta)\} \\ &v_C=\{0,V\}\end{align}

so for a given $t=T$

\begin{align*}&p_A=\{UT,0\} \\ &p_B=\{W \cos(\theta)T,W \sin(\theta)T\} \\ &p_C=\{0,VT\}\end{align}

now find the equation of the line determined by $p_A and p_C$ and show that your constraint equation leads to B lying on this line.

4. ## Re: Relative Velocity

you give me information

can you find the equation of the line that connects particle A and particle C ? If not, why not?

5. ## Re: Relative Velocity

yes i find out the equation,then what can i do>>>>

6. ## Re: Relative Velocity

show that $p_B$ is on this line given it's coordinates at time T that I showed a couple posts ago. You'll see the condition that B is on this line is identical to the equation you are given relating the velocities of the 3 particles.