Question

An aeroplane is flying vertically upwards with a uniform speed of 500 meter per second. When it is at a height of 11000 meter above the ground a shot is fired at it with the speed of 700 meter per second from a point directly below it. The minimum uniform acceleration of the aeroplane now so that it may escape from being hit?
$\left( A \right)10 \text{meter}\ \text{per}\ {\text{second}^2}$
$\left( B \right)8 \text{meter}\ \text{per}\ {\text{second}^2}$
$\left( C \right)12 \text{meter}\ \text{per}\ {\text{second}^2}$
$\left( D \right)$ None of these

Answer 1

Hint – We will start solving this question by noting down the given information and with the help of the equations of motion, i.e., $v = u + at$,$s = ut + \dfrac{1}{2}a{t^2}$ and ${v^2} - {u^2} = 2as$, where $u$ and $v$ are initial and final velocities, $s$ is the distance but here we will use height $h$ above the ground, $t$ is time and $a$ is the acceleration.
Formula used - $v = u + at$, $h = ut + \dfrac{1}{2}a{t^2}$and ${v^2} - {u^2} = 2as$

Complete step-by-step solution -
Now, given that,
Uniform speed of the aeroplane = 500 meter per second
Height of the aeroplane above the ground = 11000 $m$
Speed of the shot = 700 $m{s^{-1}}$
Now, there is a situation in which the aeroplane survives the hit if the speed of the shot becomes 500 meter per second from 700 meter per second, in this situation the probability of being hit by the shot will become 0.
And the time when the speed becomes 500 meter per second be,
$v = u + gt$
We have, $u$= 700 meter per second, $v$= 500 meter per second, $g$= 10 meter per second (which is the value of acceleration due to gravity).
$\Rightarrow 500 = 700 - 10 \times t$
$\Rightarrow 10t = 700 - 500 \\\ \Rightarrow 10t = 200 \\\ \Rightarrow t = \dfrac{{200}}{{10}} \\\ \Rightarrow t = 20\sec . \\\$
And in this time the distance travelled up by the shot is,
$h = ut + \dfrac{1}{2}g{t^2}$
\Rightarrow h = 700 \times 20 - \dfrac{1}{2} \times 10 \times {\left( {20} \right)^2} \\\ \Rightarrow h = 14000 - 5 \times 400 \\\ \Rightarrow h = 14000 - 2000 \\\ \Rightarrow h = 12000m \\\
In the same situation the distance travelled by the plane is 12000$m$- 1000 $m$= 11000 $m$.
Velocity of shot with respect to plane = 700 – 5000 = 200 meter per second
Let the acceleration of the plane be $a$.
As the bullet is going upward so acceleration of the bullet is going downwards.
So, acceleration of bullet with respect to plane = $- g - a$= $- 10 - a$
The shot will not hit the aeroplane when its velocity is minimum, i.e., $v = 0$
So, from ${v^2} - {u^2} = 2as$, we get
$\Rightarrow {0^2} - {\left( {200} \right)^2} = 2\left( { - 10 - a} \right)1000$
$\Rightarrow 0 = 40000 + 2000\left( { - 10 - a} \right)$
$\Rightarrow a = 10 \text{meter}\ \text{per}\ {\text{second}^2}$
Hence, option A is the right answer.

Note – An object is said to be moving with uniform speed, if it covers equal distance in equal intervals of time, however small these time intervals may be. The acceleration of an object is said to be uniform acceleration if its velocity changes by equal amounts in equal intervals of time, however small these time intervals may be.