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(A) Given:Horizontal velocity of the plane,$u = 360 	ext{ km/h} = 360 	imes \frac{5}{18} 	ext{ m/s} = 100 	ext{ m/s}$.Altitude (vertical height),$

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$1000 	ext{ m}$

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(A) Given:Horizontal velocity of the plane,$u = 360 	ext{ km/h} = 360 	imes \frac{5}{18} 	ext{ m/s} = 100 	ext{ m/s}$.Altitude (vertical height),$h = 500 	ext{ m}$.Acceleration due to gravity,$g = 10 	ext{ m/s}^2$.When the bomb is dropped,it follows a projectile path. The time taken to reach the ground is given by $t = \sqrt{\frac{2h}{g}}$.$t = \sqrt{\frac{2 	imes 500}{10}} = \sqrt{100} = 10 	ext{ s}$.The horizontal distance covered by the bomb during this time is $R = u 	imes t$.$R = 100 	ext{ m/s} 	imes 10 	ext{ s} = 1000 	ext{ m}$.Therefore,the bomb should be dropped at a distance of $1000 	ext{ m}$ ahead of the target.

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