A helicopter on flood relief mission, flying horizontally with a speed u at an altitude H, has to driop a food packet for a victim standing on the ground. At what distance from the victim should the packet be dropped? The victim stands in the vertical plane of the helicopters's motion.

Explanation

The velocity of the food packet at the time of rrelease is u and is horizontal. The vertical velocity at the time of release is zero
Vertical motion: If t be the time taken by the packet to reach the victim, we have for vertical motion,
`H= 1/2 gt^2 or, t= sqrt((2H)/g)` ........i.
Horizontal motion: If D be the horizontal distance travelled by the packet, we have D= ut. Putting t from i,
`D=sqrt((2H)/g)`
The distance between the victim and the packet at the time to release is
`sqrt(D^2+H^2)=sqrt((2u^2H)/g+H^2)`