Baseball Math?

Question

An outfielder picks up a ground ball and wants to make a quick 150ft throw to first base. To save time, he does not throw the ball directly to the first baseman. He skips the ball off the artificial surface at 75ft so that the ball reaches first base after a bounce. The equations

x = v_o (t)(cosΘ)
and
y = -16t^2 + v_o (t)(sinΘ) + h_o

give the coordinates of the ball at time t seconds for an initial angle of Θ degrees, initial velocity v_o ft/sec, and initial height of 5 ft and the ball is caught at a height of 5 ft.

(1)At what angle from horizontal must the ball be thrown to reach first base without a skip?
(2)At what angle from horizontal must the ball be thrown so that is strikes the ground at 75ft?

They are measured in Feet, and, yes the x-coordinate is the distance of the baseball in feet from the catcher, the y-coordinate is the height of the baseball at t seconds. I made a diagram, but photobucket is sucking right now.

pitch____________catch distance between is 150ft. point of skip is 75ft. velocity is 130ft/sec.

http://img408.imageshack.us/img408/5324/baseballdc9.png

Answer 1

x = v*t*cosθ
y = -16t^2 + v*t*sinθ + h

1) We desire y = h when x = 150
y = h means -16t^2 + vtsinθ = 0
t = v*sinθ /16

x = v^2*sinθcosθ / 16
= v^2 * sin2θ / 32 = 150
With v = 130, sin2θ = 0.284
θ = 8.25 degrees

2) Now we desire x = 75 when y = 0
x = 75 means t = 75 / (vcosθ)
Sub in y eqn
y = 0 = -900/169 * sec^2 θ + 75tanθ + 5
= -900/169 tan^2 θ + 75tanθ + 5 - 900/169

Solving, tanθ = 4.34 x 10^-3, 14.079
θ = 0.25 , 85.9 degrees
Only the first one makes sense in this problem

Answer 2

How exactly are these coordiantes measured? Is one variable the height and the other the distance between the outfielder and first baseman?