Answer
Speed range is, $3\sqrt{gD}\leq v_0 \leq 3.13\sqrt{gD}$
Work Step by Step
Using equation of trajectory,
$y=xtan\theta-\frac{gx^2}{2u^2cos^2\theta}$
For minimum possible speed, at $x=6D \Rightarrow y=2D$
So, $2D=6Dtan45^0-\frac{g(6D)^2}{2v_0^2cos^245^0} \Rightarrow 4D= \frac{36gD^2}{v_0^2} \Rightarrow v_{0,min}=3\sqrt{gD}$
Similarly, for maximum possible speed, at $x=7D \Rightarrow y=2D$
So, $2D=7Dtan45^0-\frac{g(7D)^2}{2v_0^2cos^245^0} \Rightarrow 5D= \frac{49gD^2}{v_0^2} \Rightarrow v_{0,max}=7\sqrt{\frac{gD}{5}}\approx3.13\sqrt{gD}$
So, speed range is, $3\sqrt{gD}\leq v_0 \leq 3.13\sqrt{gD}$
