Answer:
As \[h=\frac{1}{2}g{{t}^{2}}\] \[\therefore \] \[t=\sqrt{\frac{2h}{g}}\] \[\text{Speed =}\frac{\text{Distance}}{\text{Time}}\] \[\sqrt{2gh}=\frac{x}{\sqrt{2h/g}}\] or \[x=\sqrt{2gh}\times \sqrt{\frac{2h}{g}}=\mathbf{2h}\]
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