A bomber is flying at a height of 9 km at 800 km/h. How far in front of the target should he drop his bombs?

The distance in front of the target is 9.524 km

Explanation:

Given;

height of the bomber above the ground, h = 9 km = 9000 m

velocity of the bomber, v = 800 km/h = 222.22 m/s

The time the bomb will get to the target is given by;

The lead distance or distance in front of the target is given by;

x = vt

x = (222.22 m/s) x (42.857 s)

x = 9523.68 m

x = 9.524 km

Therefore, the distance in front of the target is 9.524 km

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