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A helicopter is flying in a straight line over a level field at a constant speed of 21.7 m/s and at a constant altitude of 11.5...Asked by Anonymous
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?
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Answered by
MathMate
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<sup>-1</sup>(v/h)
=tan<sup>-1</sup>(v/5.8)
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<sup>-1</sup>(v/h)
=tan<sup>-1</sup>(v/5.8)