[ vật lí 10]Rơi tự do

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nhungkyuiuep

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levietdung1998

\[\begin{array}{l}
a/ \to S = \frac{1}{2}g{t^2} \to t = \sqrt {\frac{{2S}}{g}} = \sqrt {0,2} \,\,\,\left( s \right)\\
b/{H_{\max }} = \frac{1}{2}g{.10^2} = 500m\\
H = 499m \to {t_1} = \sqrt {99,8} \,\,\,\,\,\,\left( s \right)\\
\to t = 10 - {t_1} = 0,01\,\,\,\,\left( s \right)
\end{array}\]
 
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