A bomb is projected with 200 , m/s at an angl | vedclass

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(D) Initial velocity of the bomb is $u = 200 \, m/s$ at $	heta = 60^o$. The horizontal component of velocity at the highest point is $v_x = u \cos(60

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$300 \, m/s$ in the horizontal direction

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(D) Initial velocity of the bomb is $u = 200 \, m/s$ at $	heta = 60^o$. The horizontal component of velocity at the highest point is $v_x = u \cos(60^o) = 200 	imes 0.5 = 100 \, m/s$. At the highest point,the vertical component of velocity is $0$. Let the mass of the bomb be $3m$. After the explosion,the three particles each have mass $m$. By the Law of Conservation of Linear Momentum in the horizontal direction: $(3m)v_x = m(v_{1x}) + m(v_{2x}) + m(v_{3x})$. Since the first and second particles move vertically,their horizontal velocity components are $v_{1x} = 0$ and $v_{2x} = 0$. Thus,$(3m)(100) = m(0) + m(0) + m(v_{3x})$. $300m = m(v_{3x}) \implies v_{3x} = 300 \, m/s$. By the Law of Conservation of Linear Momentum in the vertical direction: $(3m)(0) = m(v_{1y}) + m(v_{2y}) + m(v_{3y})$. Given $v_{1y} = 100 \, m/s$ and $v_{2y} = -100 \, m/s$. $0 = m(100) + m(-100) + m(v_{3y}) \implies v_{3y} = 0$. Therefore,the third particle moves with a velocity of $300 \, m/s$ in the horizontal direction.

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