# a ball is dropped from the top of a building if it covers second half of a building in 1 second. Find out height of the building

$lettheheightbeH\phantom{\rule{0ex}{0ex}}letthevelocityatsecondhalfbeu\phantom{\rule{0ex}{0ex}}\frac{H}{2}=u+\frac{g}{2}\phantom{\rule{0ex}{0ex}}{u}^{2}=gH\phantom{\rule{0ex}{0ex}}u=\sqrt{gH}\phantom{\rule{0ex}{0ex}}\frac{H}{2}=\sqrt{gH}+\frac{g}{2}\phantom{\rule{0ex}{0ex}}H=2\sqrt{gH}+g\phantom{\rule{0ex}{0ex}}solvetheaboveeqntogetHputg=10m/{s}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}regards$

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