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

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(D) The initial velocity of the bomb is $u = 200\,m/s$ at $	heta = 60^o$. At the highest point,the vertical component of velocity is $0$,and the hori

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

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(D) The initial velocity of the bomb is $u = 200\,m/s$ at $	heta = 60^o$. At the highest point,the vertical component of velocity is $0$,and the horizontal component is $v_x = u \cos(60^o) = 200 	imes 0.5 = 100\,m/s$.The total momentum at the highest point is $P = M v_x = M(100)$,where $M$ is the total mass.After the explosion,the mass splits into three equal parts $m = M/3$. Let the velocities of the three particles be $\vec{v}_1, \vec{v}_2, \vec{v}_3$.Given $\vec{v}_1 = 100\hat{j}$ and $\vec{v}_2 = -100\hat{j}$.By the Conservation of Linear Momentum in the horizontal direction:$M(100) = m(v_{1x}) + m(v_{2x}) + m(v_{3x})$$M(100) = (M/3)(0) + (M/3)(0) + (M/3)(v_{3x})$$100 = v_{3x} / 3 \implies v_{3x} = 300\,m/s$.Since the vertical components of the first two particles cancel out,the third particle must have a vertical velocity of $0$ to conserve momentum in the vertical direction.Thus,the velocity of the third particle is $300\,m/s$ in the horizontal direction.

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