An aero plane is flying vertically upwards. When it is at a height of $1000 m$ above the ground and moving at a speed of $367 m / s$ , a shot is fired at it with a speed of $567 m / s$ from a point directly below it. What should be the acceleration of aero plane so that it may escape from being hit?

> 10 m / s^{2}

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Not possible

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< 10 m / s^{2}

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> 5 m / s^{2}

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Solution

The correct option is A $> 10 m / s^{2}$

Given,
$v_{A} = 367 m / s$
$v_{s} = 567 m / s$

Then relative acceleration
$v_{r e l} = v_{p} - v_{s} = - 200 m / s$

Now let acceleration of plane $= a_{p}$
And acceleration of shot $= a_{s} = - g$
Then
$a_{r e l} = a_{p s} = a_{p} - a_{s}$
$a_{r e l} = a_{p} + g$

From Newton’s third equation of motion:

$v_{r e l}^{2} = u_{r e l}^{2} + 2 a_{r e l} s$

$0 - (200)^{2} = 2 (a_{r e l}) (- 1000)$

$a_{r e l} = 20 m / s^{2}$

So, to avoid the hit, take $g = 10 m / s^{2}$

$a_{r e l} > 20 m / s^{2}$

$a_{p} > 10 m / s^{2}$

Correct option is (b).

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