A projectile is fired with velocity u at an a | vedclass

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(D) At the highest point,the velocity of the projectile is $v_x = u \cos 	heta$ and $v_y = 0$. The total mass is $M = m + m + 2m = 4m$.Since there ar

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$\frac{5}{2} u \cos 	heta$

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(D) At the highest point,the velocity of the projectile is $v_x = u \cos 	heta$ and $v_y = 0$. The total mass is $M = m + m + 2m = 4m$.Since there are no external horizontal forces during the explosion,linear momentum is conserved along the $x$-axis.Initial momentum along $x$-axis: $P_{ix} = (4m)(u \cos 	heta)$.After the explosion:- The first part $(m)$ falls vertically,so its horizontal velocity is $0$.- The second part $(m)$ returns to the point of projection,meaning it must have a horizontal velocity of $-u \cos 	heta$.- Let the velocity of the third part $(2m)$ be $V$.Applying conservation of momentum along $x$-axis:$4m(u \cos 	heta) = m(0) + m(-u \cos 	heta) + 2m(V)$$4m u \cos 	heta = -m u \cos 	heta + 2m V$$5m u \cos 	heta = 2m V$$V = \frac{5}{2} u \cos 	heta$.

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