An enemy plane is flying horizontally at an altitude of 2 km with a speed of 300 ms^-1. An army man with an anti-aircraft gun on the ground sights the enemy plane when it is directly overhead and fires a shell with a muzzle speed of 600 ms^-1 at angle 60⁰ with the horizontal. At what minimum altitude should the enemy plane fly to avoid being hit? Take g = 10 ms^-2.

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12.5 km

13.5 km

14.5 km

15.5 km

answer is B.

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Detailed Solution

To avoid being hit, the plane should have a minimum altitude = maximum height attained by the shell which is

$h_{m a x} = \frac{v_{0}^{2} \sin^{2} θ}{2 g} = \frac{{(600)}^{2} \times \sin^{2} (60 °)}{2 \times 10}$

= 13.5 x 10³ m = 13.5 km.

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