Shouldn't this be -36.9 degrees below the horizontal? The inverse tangent shown in 4.4.32 has the x-velocity on top of the y-velocity, while the inverse tangent should actually have the y-velocity on top.

Also, it wouldn't be possible for the velocity to be 53.1 degrees below the horizontal before dropping below the point it was released, as the ball left at 45 degrees above the horizontal, and would reach the same y-position as it had before it left with the same angle below the horizontal, so the ball would have to fall below the point it was released to reach 53.1 degrees below the horizontal.

10^14 m^2 = 10^8 km^2

There's an extra 1 on the (1.68110^-27 kg) tern, it should be (1.6810^-27 kg)