# Acceleration along with Relative motion

I came across following problem:

An Aeroplane is flying vertically upwards with a uniform speed of $500m/s$. When it is at a height of $1000m$ above the ground a shot is fired at it with a speed of $700m/s$ from a point directly below it. The minimum uniform acceleration of the areoplane now so that it may escape from being hit? $(g=10m/s^2$)

My thoughts:

Relative velocity of shell shotted w.r.t. aeroplane =$700-500=200m/s$ if aeroplane if at rest (assume).

For shell to reach plane, time (t)=

$\begin{cases}a&=-10m/s^2\\ u&=200m/s\\ h&=1000m \end{cases} \implies 1000=200t-5t^2\implies t^2-40t+200=0\implies t=10(2\pm \sqrt{2})$

But how two different times for same object to cover same distance with same $u$ and $a$?

Suppose shell reaches plane in time=t sec, then plane should acquire the $a$ such that it speed becomes that of shell, so that relative distance between shell and plane does not change, then shell will not hit plane, is this thinking correct?

You are solving w.r.t plane but after the bullet is fired the plane starts accelerating . So if this acceleration is x the acceleration of bullet w.r.t plane will be g+x this is where you have gone wrong . The acceleration is not 10 it is 10+x. h=1000, u=200. For minimum accelration the we have to solve for case when bullet just hits plane that it hits when bullet is at its max height. So, $$v^2=u^2+2as$$ V=0, for max height

$$0^2=200^2+2(1000)×(-(10+x))$$

$$40000=2000(10+x)$$

$$20=10+x$$

$$x=10$$

• Did you get it? – ATHARVA Jun 1 '17 at 13:05
• is should be $x>10$ ? – mathlover Jun 2 '17 at 2:59
• yes of course or bullet wont hit plane as the relative acceleration is g-a – ATHARVA Jun 2 '17 at 9:07
• but they wrote answer directly as $10$. – mathlover Jun 2 '17 at 12:24
• Oh sorry I didn't read your comment correctly.and i have done a big mistake while writing the above ans. I will post the complete solution. – ATHARVA Jun 2 '17 at 12:44