A plane is flying horizontally at a height of 1960 m at `600 km*h^(-1)` with respect to the ground . On reaching a point directly above A , the plane drops an object which reaches the ground at B. Find the distance AB.

Explanation

Let the point directly above A from where the object is dropped be O [Fig.2.66].

Therefore , OA=1960 m . Let the time taken by the object to hit the ground at B be t.
Since the plane is flying horizontally , the initial vertical velocity of the object =0 and vertical (downward) displacement =1960 m.
Considering the vertical motion of the object , we get from the equation `h=ut+1/2 "gt"^2, `
`1960 =0xxt+1/2 xx9.8 xxt^2`
`or, t^2=(2xx1960)/(9.8) =400 or, t=20 s`
The initial horizontal velocity of the body
`=600km*h^(-1) =(600xx1000)/(60xx60)m*s^(-1) =(500)/(3) m*s^(-1)`
So, the object moves at a uniform velocity of `(500)/(3) m*s^(-1)` in the horizontal direction . Hence,
`AB=(500)/(3)xx20m=(500xx20)/(3xx1000)km=(10)/(3) =3.33km`.