Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
A baseball is thrown at an angle of 40.0° above the horizontal. The horizontal component of the baseball’s initial velocity is...Asked 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?
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?
Answers
Answered by
bobpursley
V(cos(theta))=12
V=12/sec40deg
V=12/sec40deg
Answered by
bobpursley
oops V=12/cos40deg
Answered by
Sunny
isn't the degree of 40 already given though? Why don't you do 12sin40?
Answered by
bobpursley
sine would give you the vertical compnent, horizontal component is cosine times velocity
There are no AI answers yet. The ability to request AI answers is coming soon!