Two particles A and B are moving in a horizon | vedclass

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(B) For particle $A$ to catch particle $B$,the component of velocity of $A$ in the direction of $B$'s motion must be equal to the velocity of $B$.Let

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$5$

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(B) For particle $A$ to catch particle $B$,the component of velocity of $A$ in the direction of $B$'s motion must be equal to the velocity of $B$.Let the velocity of $A$ be $v$. The velocity of $B$ is $30 \ m/s$ in the vertical direction.For $A$ to catch $B$,the vertical component of $A$'s velocity must equal $B$'s velocity:$(v_y)_A = (v_y)_B$$v \sin 37^{\circ} = 30 \ m/s$Since $\sin 37^{\circ} = 3/5$,we have:$v 	imes (3/5) = 30$$v = 30 	imes (5/3) = 50 \ m/s$Now,the time taken to catch $B$ is the time required for $A$ to cover the horizontal distance of $200 \ m$ relative to $B$ at the relative horizontal velocity:Time $t = \frac{	ext{Distance}}{	ext{Relative horizontal velocity}} = \frac{200}{v \cos 37^{\circ}}$Since $\cos 37^{\circ} = 4/5$,we have:$t = \frac{200}{50 	imes (4/5)} = \frac{200}{40} = 5 \ s$.

Two particles $A$ and $B$ are moving in a horizontal plane as shown in the figure at $t = 0$. The time after which $A$ will catch $B$ is.......$s$.

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