Answer:
\[h=\frac{1}{2}g{{t}^{2}}\] ??????..(1) \[h-20=\frac{1}{2}g{{(t-1)}^{2}}\] ??????..(2) Subtracting (1) and (2) \[\Rightarrow 20=\frac{1}{2}g\left\{ {{t}^{2}}{{(t-1)}^{2}} \right\}\] \[20=5(2t-1)\,\,or\,\,t=2.5\] \[\therefore \,\,h=\frac{1}{2}\times 10\times 6.25=31.25\,\,m\]
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