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\(T=\int_{0}^{T}dt=\int_{0}^{L}\frac{dt}{ds}ds=\int_{0}^{L}\frac{ds}{v}=\frac{1}{\sqrt{2g}}\int_{0}^{L}\frac{ds}{\sqrt{{y}_{1}-y}} =\frac{1}{\sqrt{2g}}\int_{{x}_{1}}^{{x}_{3}}\sqrt{\frac{1+{y'}^{2}}{y}}dx \)

 
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 Mar 20, 2019

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