\[(a)\;\frac{15 \sqrt 3}{2} \; m \quad (b)\;\frac{40}{3}m \quad (c)\;5 \sqrt 3\; m\quad(d)\;\frac{20}{3}\;m \]

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For a projection on an inclined plane

Time of flight $T= \large\frac{2u \sin ( \alpha-\beta)}{y \cos \beta}$

$OA=Range=\large\frac{2 u^2 \sin ( \alpha- \beta) \cos \alpha}{g \cos ^2 \beta}$

Therefore $Range= \large\frac{2 \times 10 ^2 \sin (60-30) \cos 60}{10 \times \cos ^2 30}$

$\qquad= \large\frac{2 \times 10 ^2 \times \Large\frac{1}{2} \times \Large\frac{1}{2}}{10 \times \Large\frac{3}{4}}$

$\qquad= \large\frac{20}{3} \;$$m$

Hence d is the correct answer.

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