Let tensions at lowest and highest points be $T_b \text{ and } T_t$ respectively.

Given: $\frac{T_b}{T_t} = \frac{5}{1}$

Write centripetal force equations

• At the lowest point: $T_b - mg = \frac{mv_b^2}{r}$ $T_b = \frac{mv_b^2}{r} + mg$

• At the highest point: $T_t + mg = \frac{mv_t^2}{r}$ $T_t = \frac{mv_t^2}{r} - mg$

Use energy conservation between bottom and top

The bob rises by height $2r$, so: $\frac{1}{2}mv_b^2 = \frac{1}{2}mv_t^2 + mg(2r)$ $v_b^2 = v_t^2 + 4gr$

Given $r = 1\,\text{m}$ and $g = 10\,\text{m/s}^2$: $v_b^2 = v_t^2 + 40$

Substitute into tension ratio

$\frac{T_b}{T_t} = 5$

Using expressions for tensions: $\frac{\frac{mv_b^2}{r} + mg}{\frac{mv_t^2}{r} - mg} = 5$

Cancel $m$ and put $r=1$: $\frac{v_b^2 + g}{v_t^2 - g} = 5$

Since $g=10$ and $v_b^2 = v_t^2 + 40$: $\frac{v_t^2 + 40 + 10}{v_t^2 - 10} = 5$ $\frac{v_t^2 + 50}{v_t^2 - 10} = 5$

Solve for $v_t$

$v_t^2 + 50 = 5(v_t^2 - 10)$ $v_t^2 + 50 = 5v_t^2 - 50$ $100 = 4v_t^2$ $v_t^2 = 25$ $v_t = 5\,\text{m/s}$

Final answer

The velocity of the bob at the highest position is: $\boxed{5\,\text{m/s}}$