During takeoff, an airplane climbs with a speed of 163 m/s at 31° above the horizontal. The speed+direction of the airplane constitute a vector quantity called velocity. The sun shines overhead. How fast is the shadow of the plane moving along the ground?

1 Answer
Jake M.
Jan 17, 2018

#cos(31^o) * 163# m/s = #139.7# m/s

Explanation:

You can look at the vector as having a vertical and a horizontal velocity which have a vector sum of magnitude #163# m/s at angle #31^o#.

We can use trigonometry to see that the horizontal speed is the cosine of the angle times the total speed. The horizontal speed of the airplane is the horizontal speed of the shadow.

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