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(C) Let the mass of the projectile be $m$ and its velocity at height $h = 490\, m$ be $v_0 = 200\, ms^{-1}$.According to the law of conservation of li

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$10$

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(C) Let the mass of the projectile be $m$ and its velocity at height $h = 490\, m$ be $v_0 = 200\, ms^{-1}$.According to the law of conservation of linear momentum,the momentum before the explosion equals the sum of the momenta of the two parts after the explosion.$m v_0 = \frac{m}{2} v_1 + \frac{m}{2} v_2$Given $v_0 = 200\, ms^{-1}$ (upwards) and $v_1 = 400\, ms^{-1}$ (upwards).$m(200) = \frac{m}{2}(400) + \frac{m}{2} v_2$$200 = 200 + \frac{1}{2} v_2$$0 = \frac{1}{2} v_2 \Rightarrow v_2 = 0\, ms^{-1}$.The second part is at rest at a height of $490\, m$.Using the equation of motion $h = ut + \frac{1}{2}gt^2$ for the second part:$490 = 0 \cdot t + \frac{1}{2} 	imes 9.8 	imes t^2$$490 = 4.9 t^2$$t^2 = \frac{490}{4.9} = 100$$t = 10\, s$.

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