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)`

Answer

`6.568 xx 10^(5)` .

Explanation

Here, `upsilon = 900 km//h =(900 xx1000)/(60xx60) = 250m//s`
`r = 2000 m,m =16000kg, F =` ?
At the lowest point centripetal force `(m upsilon^(2) //r)` is provided by the net force `(F -mg)` towards the centre of the circle
`:. F - mg = (m upsilon^(2))/(r)`
`F = m (g + upsilon^(2)//r) = 16000 (9.8 + (250 xx250)/(2000))`
`F = 6.568 xx 10^(5) N` .