I've shown my work in case the student would like to know where the numbers come from.

The projectile has a muzzle velocity r of 700 m/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 m/sec.

The projectile will describe a parabolic ballistic where its maximum elevation will be such that it will yield a vertical velocity of 0 at the peak. The acceleration of gravity is 9.8 m/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 / 9.8 = 50.5 seconds. Double it (after it hits apogee it has to fall back down to earth) so it spends a total of 101 seconds in flight. The trips up and down are symmetrical in space and time. Using the distance formula, it reaches an altitude of 12.496 km.

Since it spends a total of 101 seconds in flight, and its x velocity is a constant 495 m/sec it travels a distance of 49.995 km.