\(T+m g=\frac{m v^2}{R}\) (at the highest point)

\(14=\frac{1\left(v^2\right)}{R(1)}+10\)

\(v^2=4 \quad \Rightarrow v=2 \,m / s\)

Using mechanical energy conservation

\(\frac{1}{2}(1) u^2=\frac{1}{2}(1)\left(2^2\right)+1(10)(2)\)

\(u^2=64 \quad \Rightarrow u=8 \,m / s\)