Problem: The horizontal component of the velocity of a projectile is 17 meters per second. The vertical component is 20 meters per second. Using a sketch, find the angle of the motion of the object with horizontal.
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Solution: Sketch a triangle to depict 1 second's worth of motion, which will be 17 meters horizontally and 20 meters vertically. The legs of the triangle represent 17 and 20 meters, and the hypotenuse represents the displacement in 1 second, which indicates the speed of motion. The angle of the hypotenuse with horizontal is arcTan( 20 / 17) = 49.62 degrees. This is the direction of motion.
Generalized Response: The x component vx of the velocity can be represented by a vector which can be thought of as depicting 1 second's worth of motion in the x direction. The y component vy can be similarly represented. Then the result of traveling for 1 second can be seen as a displacement in the x direction followed by a displacement in the y direction. The direction of this net displacement is clearly the direction of motion, and the magnitude of this displacement indicates the speed of the object.
We can therefore regard the horizontal and vertical speeds as velocity components vx and vy of a velocity vector. The magnitude of the velocity vector will be `sqrt(vx^2 + vy^2) and its direction as measured from the positive x direction will be arctan(vy / vx).
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Figure description: The figure below depicts the velocity components vx and vy and the vector velocity resulting from the combination of these horizontal and vertical velocities. The resultant velocity is depicted by the vector comprising the hypotenuse of the triangle whose legs are vx and vy.
