During takeoff, an airplane climbs with a speed of 180m/s at an angle of 42 degrees above the horizontal. The speed and direction of the airplane constitute a vector quantity known as the velocity. The sun is shining overhead. How fast is the shadow of the plane moving along the ground? (That is, what is the magnitude of the horizontal component of the plane's velocity?)

asked by Mel

8 years ago

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1 answer

v upwards=180*sin42
v horizon=180*cos42

answered by bobpursley

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Question

During takeoff, an airplane climbs with a speed of 180m/s at an angle of 42 degrees above the horizontal. The speed and direction of the airplane constitute a vector quantity known as the velocity. The sun is shining overhead. How fast is the shadow of the plane moving along the ground? (That is, what is the magnitude of the horizontal component of the plane's velocity?)

1 answer

v upwards=180*sin42
v horizon=180*cos42

Ask a New Question or answer this question.

bobpursley · Accepted Answer

v upwards=180*sin42 v horizon=180*cos42