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onyebuchinnaji

Answer:

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;

[tex]t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2*9000}{9.8} }\\\\ t = 42.857 \ s[/tex]

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

Cricetus

In front of the target, the distance will be "9.524 km".

Distance and Height

According to the question,

Bomber's height, h = 9 km or,

                                = 9000 m

Bomber's velocity, v = 800 km/h or,

                                  = 222.22 m/s

Now,

The time will be:

→ t = [tex]\sqrt{\frac{2h}{g} }[/tex]

By substituting the values, we get

     = [tex]\sqrt{\frac{2\times 9000}{9.8} }[/tex]

     = [tex]\sqrt{\frac{18000}{9.8} }[/tex]

     = 42.857 s

hence,

The lead distance will be:

→ x = vt

By putting the values,

     = 222.22 × 42.857

     = 9523.68 or,

     = 9.524 km

Thus the above answer is correct.

Find out more information about distance here:

https://brainly.com/question/4931057

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