physics
posted by Anonymous on .
A helicopter is flying in a straight line over a level field at a constant speed of 6.2 m/s and at a constant altitude of 11.4 m. A package is ejected horizontally from the helicopter with an initial velocity of 12.0 m/s relative to the helicopter, and in a direction opposite the helicopter's motion. What angle does the velocity vector of the package make with the ground at the instant before impact, as seen from the ground?

The following calculations assume no air resistance.
Calculate the vertical component of velocity v at impact from the vertical distance S=11.4 m, a=g=9.8 m/s/s, and initial velocity u=0 (package ejected horizontally) from
v²u²=2aS
The horizontal component of velocity (h) at impact is the same as at ejection, namely 6.212=5.8.
The angle with the horizontal is tan^{1}(v/h)
=tan^{1}(v/5.8)