Question

An airplane is flying vertically upwards with a uniform speed of 50 m/s. When it is at a height of 1000m above the ground a shot is fired at it with a speed of 700 m/s from a point directly below it. The minimum uniform acceleration of the airplane so that it may escape from being hit? ($g = 10m/{s^2}$)
$
  {\text{A}}{\text{. 9}}{\text{.16 m/}}{{\text{s}}^2} \\
  {\text{B}}{\text{. 8}}{\text{.16 m/}}{{\text{s}}^2} \\
  {\text{C}}{\text{. 12 m/}}{{\text{s}}^2} \\
  {\text{D}}{\text{. None of these}} \\
$

Answer 1

Hint: The distance travelled by the airplane should be greater than the distance travelled by the gun shot. Form a quadratic equation in time of the distances travelled by the airplane and the gunshot, using the second equation of motion. The distance travelled by airplane should always be greater than that of gunshot. Impose conditions on the quadratic to get ‘a’.

Complete step-by-step answer:
Formula used:

The second equation of motion

$s = ut + \dfrac{1}{2}a{t^2}$

Let the time be ‘t’.
From the second equation of motion:

Distance travelled by the gun shot will be:
${s_g} = 700t + \dfrac{1}{2}( - 10){t^2}$

Because, g will act downwards and the bullet goes up. Opposite directions therefore the sign comes negative.

Distance travelled by the airplane will be:

${s_a} = 1000 + 500t + \dfrac{1}{2}a{t^2}$
As the gunshot should never hit the airplane so,

${s_a} > {s_g}$

$
  1000 + 500t + \dfrac{1}{2}a{t^2} > 700t - \dfrac{1}{2}(10){t^2} \\
  2000 + 1000t + a{t^2} > 1400t - (10){t^2} \\
  (a + 10){t^2} - 400t + 2000 > 0 \\
$

For such a quadratic equation solution exists when:
$D = {b^2} - 4ac < 0$
Now putting the values and checking:

$
  160000 - 4(2000)(a + 10) < 0 \\
  160000 - 8000a - 80000 < 0 \\
  80000 < 8000a \\
  10 < a \\
$

As the value of acceleration should be greater than 10, so 12 m/s² from the options satisfies the inequality.

The correct option is (C).

Note: For a quadratic inequality to be greater than 0 the value of the discriminant should be less than 0 for the quadratic to have solutions.