Not an expert on the subject, but here is how I would solve it.
Using the simple http://en.wikipedia.org/wiki/Trajectory_of_a_projectile" , you have the following equations:
http://en.wikipedia.org/wiki/Trajectory_of_a_projectile#Height_at_x":
Where
y0 = 0 and \theta = 45° (\theta is the angle at which the projectile is launched; 45° is the value that will give the greatest distance). Knowing
y (the height of the left field fence, 4ft) and
x (the distance between the player and the fence), you can find the initial velocity
v.
Using the same equation with the same
v but with
x that is 15 ft shorter, you can find the new value of
y (the height of the fence 15 ft closer for the same projectile path). Re-arranging for your problem, you get:
y_{R} = \left( x - 15 \right) - \frac{\left( x - 15 \right) ^{2}}{x^{2}} \left( x - 4 \right)
Where y_{R} is the height of the RIGHT field fence and x is the distance between the player and the LEFT field fence. Solutions for this equation are:
x = 50 ft --> y_{R} = 12.5 ft;
x = 75 ft --> y_{R} = 14.6 ft;
x = 100 ft --> y_{R} = 15.6 ft;
x = 150 ft --> y_{R} = 16.7 ft;
x = 200 ft --> y_{R} = 17.3 ft;
x = 250 ft --> y_{R} = 17.6 ft;
x = 300 ft --> y_{R} = 17.9 ft;
x = 350 ft --> y_{R} = 18.0 ft;
x = 400 ft --> y_{R} = 18.1 ft;
x = \infty --> y_{R} = 19 ft;
I did not try for other angles to see how different are the results.
