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Kinematics airplane problem

  • Thread starter jehan4141
  • Start date
91
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An airplane with a speed of 97.5 m/s is climbing upward at an angle of 50.0° with respect to the horizontal. When the plane's altitude is 732 m, the pilot releases a package.

(a)
Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.

(b)
Relative to the ground, determine the angle of the velocity vector of the package just before impact.



I got part A. The answer is ~1380 meters. However, I do no understand Part B. The answer should be 66 degrees below the ground, but I do not see how.

My Thoughts
The path of the dropped package is symmetrical. Thus, with respect to the +x axis, shouldn't the direction equal 130 degrees????
 
6
0
You know the planes x velocity is 97.5(cos50) therefore this will be the packages x velocity (ignoring air resistance). You can find the y velocity because you know the height at which the package is dropped. Use these two velocities as the sides of a right triangle and using the inverse tangent function, find that angle. This will be your answer.
 

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