A) \[2.5\sqrt{{}}3\]
B) \[62.5\sqrt{{}}3\]
C) \[125\sqrt{{}}3\]
D) \[9.65\sqrt{{}}3\times {{10}^{-3}}\]
Correct Answer: D
Solution :
: Range of projectile \[=R=\frac{{{u}^{2}}\sin 2\theta }{g}\] \[\therefore \] \[R={{\left( \frac{50\times 1000}{60\times 60} \right)}^{2}}\times \frac{\sin 60{}^\circ }{10}\] Or \[R=\frac{125\times 125\times \sqrt{3}}{9\times 9\times 2\times 10}m\] Or \[R=9.65\sqrt{3}m\] Or \[R=9.65\sqrt{3}\times {{10}^{-3}}km\]
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