Question

Two particles are projected simultaneously from the level ground which are x distance apart. They may meet after a time:

Answer 1

Given:

Two particles are projected simultaneously from the level ground which are x distance apart.

To find:

Time after which the particles will meet.

Calculation:

The particles will meet only when their x and y coordinates match ,i.e. their positions become same.

Taking the 2nd particle to be of reference frame;

x component of Velocity of 1st particle will be

$= u1 \cos( \theta1) - u2 \cos( \theta2)$

x component velocity of second particle will be

$= 0$

They are separated by x ;

$x = \{u1 \cos( \theta1) - u2 \cos( \theta2) \} \times t$

$\boxed{ = > t= \dfrac{x}{u1 \cos( \theta1) - u2 \cos( \theta2) }}$

Now , the y position should also be same for both the particles;

$u1 \sin( \theta1) t - \dfrac{1}{2} g {t}^{2} = u2 \sin( \theta2) - \dfrac{1}{2} g {t}^{2}$

$= > u1 \sin( \theta1) t = u2 \sin( \theta2) t$

$= > u1 \sin( \theta1) = u2 \sin( \theta2)$

$\boxed{ = > u2 = \dfrac{u1 \sin( \theta1) }{ \sin( \theta2) }}$

Putting value of u2 in 1st Equation;

$\therefore \: t= \dfrac{x}{u1 \cos( \theta1) - u2 \cos( \theta2) }$

$= > \: t= \dfrac{x}{u1 \cos( \theta1) - \bigg \{ \frac{u1 \sin( \theta1) }{ \sin( \theta2) } \bigg \} \cos( \theta2) }$

$= > \: t= \dfrac{x \sin( \theta2) }{u1 \bigg \{ \cos( \theta1) \sin( \theta2) - \sin( \theta1) \cos( \theta2) \bigg \}}$

$= > \: t= \dfrac{x \sin( \theta2) }{u1 \bigg \{ \sin( \theta2 - \theta1) \bigg \}}$

So, final answer is:

$\boxed{ \red{ \bold{\: t= \dfrac{x \sin( \theta2) }{u1 \bigg \{ \sin( \theta2 - \theta1) \bigg \} }}}}$