Question

An Aeroplane in flying vertically upward. When it is at a height of $1000m$ above the ground and moving at a speed of $367m/s$, a shot is fired at it with a speed of $567m{s}^{-1}$ from a point directly below it. What should be the acceleration of Aeroplane so that it may escape from being hit?

• A. $>5m{s}^{-2}$
• B. $>10m{s}^{-2}$
• C. $<10m{s}^{-2}$
• D. Not possible

Answer 1

Answer: (B)

$>10m{s}^{-2}$

The time in which, the speed of shot becomes $367m/s$, can be written as,

$v=u-gt$

$367=567-10\times t$

$t=20\sec$

The distance travelled by the shot is given as,

$h=ut-\frac{1}{2}g{t}^2$

$h=567\times 20-\frac{1}{2}\times 10\times {\left(20\right)}^2$

$h=9340m$

The total distance travelled by the plane is given as,

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

$9340=367\times 20+\frac{1}{2}a\times {\left(20\right)}^2$

$a=10m/s^2$

Thus, the acceleration of airplane is $10m/s^2$.

So the acceleration must be greater than $10m/s^2$