Lời giải

Ta có: \({v_{max}} = \sqrt {2g\ell \left( {1 - \cos {5^0}} \right)} \)

\(s_0^2 = \frac{{v_{max}^2}}{{{\omega ^2}}} = \frac{{2g\ell \left( {1 - \cos {5^0}} \right)}}{{\frac{g}{{\frac{\ell }{2}}}}} \Rightarrow {\alpha _0} = 0,123\left( {rad} \right) = {7,1^0}\)