1
2
4
1/2

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This problem is related to the value of the velocity components in a given point. Since the vertical displacement of the projectile can be modeled by:

$y=\frac{gt^2}{2} \rightarrow t=\sqrt{\frac{2y}{g}}$

So, the magnitude of the velocity's components at the landing point is given by:

$v_x=d \sqrt{\frac{g}{2h}} ; v_y=\sqrt{2gh}$

At the landing point, the velocity vector makes an angle of 45º below the ground, meaning that the magnitude of the velocity components should be the same:

$d \sqrt{\frac{g}{2h}}=\sqrt{2gh} \rightarrow \frac{d}{h}=2$

Thus, the anwser is 2.