Question

what must be the the minimum initial velocity to reach 1000m range? what is the total time required to reach this range

Answer 1

Given:

$R=1000\:\rm m$

The range

$$R=\frac{v_0^2\sin2\theta}{g}$$

The minimum speed corresponds to the angle of throws $\theta=45^{\circ}$ and equal

$$v_0=\sqrt{gR}=\sqrt{9.8*1000}=99\:\rm m/s$$