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(C) Let the mass of the projectile be $2m$. At the highest point,its velocity is horizontal,$v_x = 20 \, m/s$. By the law of conservation of linear mo

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(C) Let the mass of the projectile be $2m$. At the highest point,its velocity is horizontal,$v_x = 20 \, m/s$. By the law of conservation of linear momentum,the initial momentum equals the final momentum. Initial momentum: $P_i = (2m)(20) \hat{i} = 40m \hat{i}$. Let the velocity of the first part (mass $m$) be $\vec{v}_1 = 30 \hat{j} \, m/s$. Let the velocity of the second part (mass $m$) be $\vec{v}_2 = v_x \hat{i} + v_y \hat{j}$. Conservation of momentum: $40m \hat{i} = m(30 \hat{j}) + m(v_x \hat{i} + v_y \hat{j})$. Equating components: Horizontal: $40 = v_x \implies v_x = 40 \, m/s$. Vertical: $0 = 30 + v_y \implies v_y = -30 \, m/s$. The velocity of the second part is $\vec{v}_2 = 40 \hat{i} - 30 \hat{j}$. The magnitude is $v_2 = \sqrt{40^2 + (-30)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, m/s$.

$A$ projectile is moving at $20 \, m/s$ at its highest point,where it breaks into two equal parts due to an internal explosion. One part moves vertically up at $30 \, m/s$. Then the other part will move at ............. $m/s$.

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