"Using physics rather than calculus" strikes me as an odd expression since the physics equations are derived from calculus, but I understand your meaning.

At the peak of its trajectory, dx/dt = 0, so we set dx/dt = c - 9.8t = 0

Solve for t, to get t = c/9.8

This is the time at which x_max is reached

plug back into equation for x(t)

x_max = ct_max - 4.9t_max^2

250 = c(c/9.8) - 4.9(c/9.8)^2

...

c = 70 m/s