A projectile of mass m is projected from grou | vedclass

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(A) $1$. Initial velocity components: $u_x = 50 \cos 53^{\circ} = 50 	imes (3/5) = 30 \, m/s$ and $u_y = 50 \sin 53^{\circ} = 50 	imes (4/5) = 40 \,

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$240$

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(A) $1$. Initial velocity components: $u_x = 50 \cos 53^{\circ} = 50 	imes (3/5) = 30 \, m/s$ and $u_y = 50 \sin 53^{\circ} = 50 	imes (4/5) = 40 \, m/s$.$2$. At the highest point,the velocity is $v_x = 30 \, m/s$ and $v_y = 0$. The horizontal distance from the origin to the highest point is $R/2 = (u_x^2 \sin 2	heta) / (2g) = (30 	imes 40) / 10 = 120 \, m$.$3$. At the highest point,the mass $m$ splits into two parts $m/2$ and $m/2$. One part comes to rest,so its velocity $v_1 = 0$. By conservation of momentum: $m v_x = (m/2) v_1 + (m/2) v_2 \implies m(30) = 0 + (m/2) v_2 \implies v_2 = 60 \, m/s$.$4$. The part at rest falls vertically to the ground at $x_1 = 120 \, m$. The second part moves horizontally with $v_2 = 60 \, m/s$ and falls from height $H = u_y^2 / (2g) = 40^2 / 20 = 80 \, m$. Time to fall $t = \sqrt{2H/g} = \sqrt{160/10} = 4 \, s$.$5$. Horizontal distance of the second part from the highest point is $d = v_2 	imes t = 60 	imes 4 = 240 \, m$. Its final position is $x_2 = 120 + 240 = 360 \, m$.$6$. The distance between the two pieces is $|x_2 - x_1| = |360 - 120| = 240 \, m$.

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