theo đề ta có
\(30=V_A+2a\)
\(80=\dfrac{V^2_A}{2a}\)
\(\Rightarrow\left\{{}\begin{matrix}2a=30-2.V_A\\2a=\dfrac{V^2_A}{80}\end{matrix}\right.\)
\(\Rightarrow V^2_A-2.V_A+30=0\)
\(\Rightarrow V^2_A-160V_A+2400=0\)
\(\Rightarrow\) Tự làm nốt...................................