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

At the lowest point in the path, the acceleration is vertically upward (towards the centre) and its magnitude is $v^2/r$.

The forces on the plane are :

(a) weight Mg downward and

(b) force F by the air upward.

Hence, Newton's second law of motion gives

$F-Mg=M{v}^2/r \text{ or } F=M\left(g+v^2/r\right)$

Here,

 $v=900 \mathrm{km}/\mathrm{h}=\frac{9\times {10}^5}{3600} \mathrm{m}/\mathrm{s}=250 \mathrm{m}/\mathrm{s}$ 

$F=16000\left(9.8+\frac{62500}{2000}\right)\mathrm{N}=6.56\times {10}^5 \mathrm{N}$