An aeroplane flying vertically upwards with a uniform speed of 500m/s. when it is at height of 1000m above ground a shot is fired at it with a speed of 700m/s from a point directly below it.The minimum uniform acceleration of aeroplane now so that it may escape from being hit?

							
For aeroplane,
from first eqn.
v = u – gt
500 = 700 – gt
t = 20 s.
and in this time,
 h = 700*20 – 0.5g*400
   = 12000m
total distamce = 12000 – 1000 = 11000 meter.
apply s = ut + 0.5at^2
for s= 11k meter and at t= 20 s and u = 500 m/s.
find a.
						

							
Velocity of bullet wrt plane = 700 -500 = 200m/s
Let the acceleration of plane be a.
As the bullet is going upward so acceleration of bullet is g downwards
So acceleration of bullet wrt plane = -g-a
                                                            = -10-a
The bullet will not hit the when its vel is min i.e v=0
So from v2 =u2+2as
0=200 2 +2(-10-a) 1000
0=40000+2000(-10-a)
a= 10m/s2
						

							
 
Velocity of bullet with respect to the plane = 700 -500 = 200m/s
Let the acceleration of plane be a.
As the bullet is going upward so acceleration of bullet is g downwards
So acceleration of bullet with respect to the plane = -g-a
                                                            = -10-a
The bullet will not hit the when its velocity is minimum that is v=0
So from v2 =u2+2as
0=200 2 +2(-10-a) 1000
0=40000+2000(-10-a)
a= 10m/s2