Đáp án:
\(\frac{{mgh}}{{\frac{1}{{\sin \alpha }}\sqrt {\frac{{2h}}{g}} }} = \frac{{\sin \alpha .mg.h}}{{\frac{{\sqrt {2h} }}{{\sqrt g }}}} = \frac{{\sin \alpha .m\sqrt {{g^2}} .{{(\sqrt h )}^2}.\sqrt g }}{{\sqrt 2 .\sqrt h }} = \frac{{\sin \alpha .m\sqrt {{g^3}} \sqrt h }}{{\sqrt 2 }} = \frac{{\sin \alpha .m\sqrt {h{g^3}} }}{{\sqrt 2 }}\)