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(D) For a projectile launched horizontally from a height $h$ with initial velocity $u = 10 \sqrt{3} 	ext{ m s}^{-1}$:$1$. The horizontal component of

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$5$

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(D) For a projectile launched horizontally from a height $h$ with initial velocity $u = 10 \sqrt{3} 	ext{ m s}^{-1}$:$1$. The horizontal component of velocity remains constant: $v_x = u = 10 \sqrt{3} 	ext{ m s}^{-1}$.$2$. The vertical component of velocity at the time of landing is $v_y = \sqrt{2gh}$.$3$. The angle $	heta$ that the velocity vector makes with the horizontal is given by $	an 	heta = \frac{v_y}{v_x}$.$4$. Given that the landing angle is $30^{\circ}$ or more,we have $	an 	heta \geq 	an 30^{\circ}$.$5$. Substituting the values: $\frac{\sqrt{2gh}}{10 \sqrt{3}} \geq \frac{1}{\sqrt{3}}$.$6$. Simplifying: $\sqrt{2gh} \geq 10$.$7$. Squaring both sides: $2gh \geq 100$.$8$. With $g = 10 	ext{ m s}^{-2}$,we get $20h \geq 100$,which implies $h \geq 5 	ext{ m}$.$9$. The minimum height is $5 	ext{ m}$.

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