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Questions and answers by
Kebede Rikitu Shogile
Answers (1)
R_{max}= v_{o}^{2}sin2theta/2g We have the maximum R if sin2theta =1 then our theta will be 45^{0} so, Now, v_{yf}^{2} = v_{y0}^{2} - 2g(y_{f} - y_{i}) y = [v_{o}^{2} - v_{yf}^{2} ) /2g] v_{yf}^{2} = 0 y = [v_{o}^{2} /2g] We have R= v_{o}^{2}/29 y = R /2
