Question

An aeroplane is flying vertically upwards with a uniform speed of $50m/s$. 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 aeroplane now so that it may escape from being hit? $(g=10m/s^2)$

• A. $9.16m/s^2$
• B. $8.16m/s^2$
• C. $12m/s^2$
• D. $Noneofthese$

Answer 1

The correct option is B $8.16m/s^2$

The shot fired $vertically$ will only be able to reach a $certain$ height, after that height it will start moving in $downward$

direction, because at that height its vertical velocity will become $zero$ fir a moment and it will reverse its direction.

This maximum height will be $h=\frac{v^2}{2g}=\frac{490000}{20}=24500meter$

time taken to reach this height or time taken to make the velocity $zero$ $momentarily$ will be $t=\frac{v}{g}=70second$

Now to escape from being hit the aeroplane need to cover $24500-1000=23500meter$ in these $70sec$

so if the uniform acceleration needed be $a$ then $23500=50\times 70+\frac{1}{2}a{70}^2=3500+2450a$

or $a=8.16m/s^2$