Posted by **Anonymous** on Friday, September 18, 2009 at 8:02pm.

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?

- physics -
**MathMate**, Friday, September 18, 2009 at 9:08pm
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

v²-u²=2aS

The horizontal component of velocity (h) at impact is the same as at ejection, namely 6.2-12=5.8.

The angle with the horizontal is tan^{-1}(v/h)

=tan^{-1}(v/5.8)

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