Answer:
The range, S = 6307.82 ft
Step-by-step explanation:
Given,
The horizontal velocity of the bomber, Vx = 800 ft/s
The altitude of the bomber, h = 1000 ft
The range of the projectile is given by the relation
S = Vx [Vy +√(Vy² + 2gh] / g
If the vertical component of the velocity is zero, the equation becomes
S = Vx √(2gh) / g
Substituting the given values,
S = 800 √(2X 32.17 X 1000) / 32.17
= 6307.82 ft
Hence, the bomb falls at a distance, S = 6307.82 ft