I say (roughly converting in my head to metric units)
1.3 km, 5 km. My reasoning, in English units.

First, the mass of the projectile is irrelevant. The mass is only a factor when we take air resistance into account, and for that we have to know the exact shape and aerodynamics of the projectile and atmospheric pressure, etc. Artillerymen look that up in tables. So let us neglect air resistance and just deal with the mechanics of the problem.

The projectile has a muzzle velocity r of 700 ft/sec and it is directed at an angle of 45 degrees. This vector can be resolved into an x-component parallel to the ground, and a y-component perpendicular to it. Use the equations x= r cos theta and y = r sin theta where r is 700 and theta = 45 degrees.

Sin theta = cos theta = 0.7071, so the horizontal and vertical components x = y = 495 ft/sec.

The projectile will describe a parabolic ballistic where its maximum elevation will be such that it will yield a velocity of 0 at the peak. The acceleration of gravity is 32.2 ft/s**2.

Velocity is acceleration times time (v = at) and distance (d = 1/2 at**2).

The vertical progress is the y velocity decreased by acceleration until it's 0, or 495 / 32.2 = 15.4 seconds. Double it (after it hits apogee it has to fall back down to earth) so it spends a total of 30.8 seconds in flight. The trips up and down are symmetrical in space and time. Using the distance formula, it reaches an altitude of 3818 feet.

Since it spends a total of 30.8 seconds in flight, and its x velocity is a constant 495 ft/sec it travels a distance of 15246'.