c
From E.C. \(=\frac{1}{2} mv _{0}^{2}= mgh\)

\(v _{0}=10 \sqrt{2}\)

For \(A \rightarrow B\)

at \(B , v =0\)

\(a =- g \sin 45^{\circ}=\frac{-10}{\sqrt{2}}\)

\(v = u + at _{1} \Rightarrow 0=10 \sqrt{2}-\frac{10}{\sqrt{2}} t _{1} \Rightarrow t _{1}=2\,sec\)

For \(B \rightarrow C\)

\(s =u t _{2}+\frac{1}{2} at _{2}^{2}\)

\(\frac{10}{\sin 30^{\circ}}=\frac{1}{2}\left(10 \sin 30^{\circ}\right) t _{2}^{2}\)

\(t _{2}=2 \sqrt{2}\)

So total time

\(T=t_{1}+t_{2}\)

\(=2 \sqrt{2}+2\)

\(=2(\sqrt{2}+1)\sec\)