There is now a constant horizontal force F
on the ball due to the wind meaning that the ball now has an acceleration in the x direction
of -F/m. The vertical motion is the same as in the previous part of the question where
we have again set y equal to 0 because we’re interested in the point where the ball lands.
Discarding the trivial t=0 solution we get t=(2u sin(theta))/g as the flight time of
the ball. Now substitute this time into the horizontal equation of motion to find the
distance the ball travels.

Tagged : # # # # # # #

Dennis Veasley

Leave a Reply

Your email address will not be published. Required fields are marked *