\(t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2\cdot60}{10}}=2\sqrt{3}s\)
Ta có: \(v^2=v_0^2+\left(gt\right)^2\)
\(\Rightarrow v_0=\sqrt{v^2-\left(gt\right)^2}=\sqrt{40^2-\left(10\cdot2\sqrt{3}\right)^2}=20\left(\dfrac{m}{s}\right)\)
Độ cao cực đại:
\(L=v_0t=20\cdot2\sqrt{3}=40\sqrt{3}m\)