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1 vote
1 vote
So I got stuck on this question

User Funcraft
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1 Answer

21 votes
21 votes

\begin{gathered} d=v0\cdot t+(at^2)/(2) \\ d=0\cdot1.6+(9.81\cdot1.6^2)/(2) \\ d=12.557m \\ \end{gathered}

The initial speed is 0, then while he falls down the speed increase

User Abdelrahman
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