Two particles A and B are projected simultane | vedclass

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Question

(B) Let the angle of projection of particle $B$ with the horizontal be $\alpha$. For particle $B$ to hit particle $A$ moving on the ground,the horizon

Vedclass · Accepted Answer

$60^{\circ}$

Vedclass · Answer

(B) Let the angle of projection of particle $B$ with the horizontal be $\alpha$. For particle $B$ to hit particle $A$ moving on the ground,the horizontal component of the velocity of $B$ must be equal to the velocity of $A$ (since $A$ moves with constant speed $v$ on the horizontal surface). Thus,$v_B \cos \alpha = v$. Given $v_B = \frac{2v}{\sqrt{3}}$,we have $\frac{2v}{\sqrt{3}} \cos \alpha = v$. $\cos \alpha = \frac{\sqrt{3}}{2}$. Therefore,$\alpha = 30^{\circ}$ with the horizontal. The question asks for the angle of projection with the vertical. Angle with vertical = $90^{\circ} - \alpha = 90^{\circ} - 30^{\circ} = 60^{\circ}$.

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