This part of the question is similar to the first part but now we must take into account the deceleration due to the wind. The horizontal equation of motion is now
x=ut cos(theta) – (F t^2)/(2m). From the vertical equation of motion we can find the time of flight t=(2u sin(theta))/g as in the previous part of the question. Substituting this into the horizontal
equation of motion gives us the range of the ball as a function of theta. Now find the angle theta which
maximizes this expression. You might find the identity sin(2 theta)=2 sin(theta) cos(theta) useful.

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Dennis Veasley

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