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Question-8667

Question Number 8667 by tawakalitu last updated on 20/Oct/16 Answered by sandy_suhendra last updated on 20/Oct/16 $$\mathrm{at}\:\mathrm{t}=\mathrm{5}\:\mathrm{sec}\:\mathrm{the}\:\mathrm{body}\:\mathrm{reaches}\:\mathrm{the}\:\mathrm{ground}\:\Rightarrow\:\mathrm{y}=\mathrm{0} \\ $$$$\mathrm{y}=\mathrm{V}_{\mathrm{o}} .\mathrm{t}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}.\mathrm{t}^{\mathrm{2}} \\ $$$$\mathrm{0}=\mathrm{V}_{\mathrm{o}} .\mathrm{5}\:−\:\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{10}.\mathrm{5}^{\mathrm{2}} \\ $$$$\mathrm{5V}_{\mathrm{o}}…