Two particles are projected simultaneously from the level ground which are x distance apart. They may meet after a time:
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Two particles are projected simultaneously from the level ground which are x distance apart. They may meet after a time:
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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
[tex] = u1 \cos( \theta1) - u2 \cos( \theta2) [/tex]
x component velocity of second particle will be
[tex] = 0[/tex]
They are separated by x ;
[tex]x = \{u1 \cos( \theta1) - u2 \cos( \theta2) \} \times t[/tex]
[tex] \boxed{ = > t= \dfrac{x}{u1 \cos( \theta1) - u2 \cos( \theta2) }}[/tex]
Now , the y position should also be same for both the particles;
[tex]u1 \sin( \theta1) t - \dfrac{1}{2} g {t}^{2} = u2 \sin( \theta2) - \dfrac{1}{2} g {t}^{2} [/tex]
[tex] = > u1 \sin( \theta1) t = u2 \sin( \theta2) t[/tex]
[tex] = > u1 \sin( \theta1) = u2 \sin( \theta2) [/tex]
[tex] \boxed{ = > u2 = \dfrac{u1 \sin( \theta1) }{ \sin( \theta2) }}[/tex]
Putting value of u2 in 1st Equation;
[tex] \therefore \: t= \dfrac{x}{u1 \cos( \theta1) - u2 \cos( \theta2) }[/tex]
[tex] = > \: t= \dfrac{x}{u1 \cos( \theta1) - \bigg \{ \frac{u1 \sin( \theta1) }{ \sin( \theta2) } \bigg \} \cos( \theta2) }[/tex]
[tex] = > \: t= \dfrac{x \sin( \theta2) }{u1 \bigg \{ \cos( \theta1) \sin( \theta2) - \sin( \theta1) \cos( \theta2) \bigg \}}[/tex]
[tex] = > \: t= \dfrac{x \sin( \theta2) }{u1 \bigg \{ \sin( \theta2 - \theta1) \bigg \}}[/tex]
So, final answer is:
[tex] \boxed{ \red{ \bold{\: t= \dfrac{x \sin( \theta2) }{u1 \bigg \{ \sin( \theta2 - \theta1) \bigg \} }}}}[/tex]