An aeroplane is flying at a constant height of 1960 m with speed `600 kmh^(-1)` above the ground towards point directly over a person struggling in flood water. At what angle of sight with the vertical should be pilot release a survival kit if it is to reach the person in water ? `(g = 9.8 ms^(-2))`

Explanation

As the plane is flying at a speed of `600xx(5//18)`, i.e., (500/3) m/s horizontally (at a height of 1960 m above the ground) the time taken by the kit to reach the ground
`t=sqrt((2h)/(g))=sqrt((2xx1960)/(9.8))`=20s
And in this time the kit will move horizontally by
`x=ut=(500//3)xx20=(10,000//3)m`
So the angle of sight
`tan phi=(x)/(h)=(10000)/(3xx1960)=(10)/(5.88)=1.7=sqrt(3)`
i.e., `phi=tan^(-1)(sqrt(3))=60^(@)`