PHYSICS HELP PLEASE!
posted by Sunny .
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?

V(cos(theta))=12
V=12/sec40deg 
oops V=12/cos40deg

isn't the degree of 40 already given though? Why don't you do 12sin40?

sine would give you the vertical compnent, horizontal component is cosine times velocity
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