Physics
EmmittLagoa838
2021-03-19 16:45:12
A baseball 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
ANSWERS
batman13245
2021-03-19 21:11:23

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

ne583
2021-03-19 21:12:38

Imagine you have a triangle at angle 40.  initial velocity= m/sec  12 is the horizontal component, so it is the cosine side, so you must find cosine(40) for the angle.  then you have to divide 12/cos(40) to get 15.7 m/s as your initial velocity

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