A fighter plane is pulling out for a dive at a speed of `900 km//h`. Assuming its path to be a vertical circle of radius `2000 m` and its mass to be `16000 kg`, find the force exerted by the air on it at the lowest point. Take `g=9.8 m//s^(2)`

Explanation

At the lowest point in the, the acceleration is vertically upward (towards the centre) and its magnitude is `v^(2)//r`.
The force on the plane are (a) Weight `Mg` downwards and
(b) force `F` by the air upward.
Hence, Newton's second law of motion gives
`F-Mg=Mv^(2)//r` or `F=M(g+v^(2)//r)`
Here, `v=900 km//h=(9xx10^(5))/3600 m//s=250 m//s`
`F=16000(9.8+62500/2000)N=6.56xx10^(5) N ` (upwards)