A projectile is fired at a speed of 100 , m/s | vedclass

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(C) Let the total mass of the projectile be $M = m + 3m = 4m$. At the highest point,the velocity is $v_x = u \cos 	heta = 100 \cos 37^\circ = 100 	i

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$1120$

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(C) Let the total mass of the projectile be $M = m + 3m = 4m$. At the highest point,the velocity is $v_x = u \cos 	heta = 100 \cos 37^\circ = 100 	imes (4/5) = 80 \, m/s$.By the Conservation of Linear Momentum $(COLM)$ in the horizontal direction:$M v_x = m(0) + 3m(v')$,where $v'$ is the velocity of the heavier part.$4m(80) = 3m(v') \Rightarrow v' = (4/3) 	imes 80 = 320/3 \, m/s$.The horizontal distance covered by the heavier part from the highest point is $x = v' 	imes t$,where $t$ is the time to fall from the highest point to the ground.$t = (u \sin 	heta) / g = (100 \sin 37^\circ) / 10 = (100 	imes 3/5) / 10 = 6 \, s$.$x = (320/3) 	imes 6 = 640 \, m$.The total distance from the launching point is $R/2 + x$,where $R$ is the range of the projectile.$R = (u^2 \sin 2	heta) / g = (100^2 	imes 2 \sin 37^\circ \cos 37^\circ) / 10 = (10000 	imes 2 	imes 0.6 	imes 0.8) / 10 = 960 \, m$.Distance $= 960/2 + 640 = 480 + 640 = 1120 \, m$.

$A$ projectile is fired at a speed of $100 \, m/s$ at an angle of $37^\circ$ above the horizontal. At the highest point,the projectile breaks into two parts of mass ratio $1:3$,the smaller coming to rest. Then the distance of the heavier part from the launching point is ............... $m$.

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