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

Plane is flying at a speed `=600xx(5)/(18)=(500)/(3) m//s` horizontaly (at a height 1960m)
time taken by the kit to reach the ground `t=sqrt((2h)/(g))=sqrt(2xx1960)/(9.8)=20s`
in this time the kit will move horizontally by `x=ut=(500)/(3)xx20=(10,000)/(3)m`
So the angle of sight `tan theta=(x)/(h_=(10,000)/(3xx1960)=(10)/(5.88)=1.7=sqrt(3)` or `theta=60^(@)`