Question

an aeroplane is flying vertically upward with a uniform speed 500m/s when it age at height of 1000 m above the ground a shot is fired at it with a speed of 700 M per second form of one directory below it the minimum acceleration of the aeroplane now so that it may escape from being hit

Answer 1

If there is a situation in which the aeroplane survives the HIT till when the speed of shot becomes 500 ms^-1 in this situation the probability of being will become 0.

And the time when the speed becomes 500 ms^-1 it will be:

v = u -gt

or 500 = 700 -10 x t

or t = 20 Seconds.

And in this time the distance traveled up by the shot, is

h = ut - 1/2 gt^2

or h = 700 x 20 - 1/2 x 10 x 400

= 1400 - 2000

h = 12000 m.

In the same situation the distance traveled by the plane is 11000 m.

then the acceleration for the air plane will be a for example.

we will use the following formula:

s = ut + 1/2 at^2

here s is the distance traveled by the airplane in time t.

s = 11000 m,

t = 20 seconds

11000 = 50 x 20 + 1/2 a x 400

11000 = 10000 + 200 a

1000 = 200 a

a = 5 ms^-1