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How do i begin to answer this question?

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A motorcycle stunt person attempts to jump over a 200 m wide canyon. He leaves one side with a velocity of 43 m/s at an angle of 40 degrees. Assuming he needs to land at the same vertical height with which he left the other side, will he make it? Support your answer with appropriate calculations.
asked Feb 16, 2014 in Highschool Physics by beean2014 (120 points)

1 Answer

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First, you need to 'separate' the velocity: along x-axis and y-axis. Therefore,

x-axis velocity is 43m/s *cos 40=32.9m/s

y-axis velocity is 43m/s *sin 40=27.6m/s

During this process, the x-axis velocity is unchanged(neglect air resistance.)

So, that guy needs 6.06s to cross the canyon.

For y-axis, consider you are throwing a ball upward as an initial v=27.6m/s and find the position after 6.06s. Set initial position is 0m. If final position is positive, he can make it. If negative, he is gonna fall into the canyon.

Hope that helps
answered Feb 16, 2014 by Will_White (1,010 points)
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