Abaseball is thrown at an angle of 40.0° above
the horizontal. the horizontal component of the
baseball’s initial velocity is 12.0 meters per
second. what is the magnitude of the ball’s initial
velocity?
(1) 7.71 m/s (3) 15.7 m/s
(2) 9.20 m/s (4) 18.7 m/s

Vector trigonometry can be used for this problem. Since the horizontal component is 12 meters per second, this is technically the hypotenuse (actual initial velocity) multiplied to cosine of 40 degrees. Therefore, to find the hypotenuse, we must divide 12 by cosine 40degrees. cos(40)= 0.766, and 12/0.766 = approximately 15.664, therefore our answer is (3) 15.7 m/s

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