Question

a bomber is flying horizontally at a speed of 800.0 ft/s and an altitude of 1000.0 ft when it is drops a bomb

Answer 1

Answer:

 The range, S = 6307.82 ft

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