1. Use conservation of mechanical energy

At the lowest position, the bob has maximum speed: $v_0 = 3\ \text{m/s}$

Length of pendulum: $l = 50\ \text{cm} = 0.5\ \text{m}$

At angle $60^\circ$ from the vertical, the bob rises by height $h$ from the lowest point.

2. Find the rise in height

For a pendulum, when displaced by angle $\theta$ from vertical, $h = l(1-\cos\theta)$

Here, $\theta = 60^\circ, \quad \cos 60^\circ = \frac{1}{2}$

So, $h = 0.5\left(1-\frac{1}{2}\right)=0.5\times \frac{1}{2}=0.25\ \text{m}$

3. Apply energy conservation

At the lowest point: $E = \frac{1}{2}mv_0^2$

At $60^\circ$: $E = \frac{1}{2}mv^2 + mgh$

Thus, $\frac{1}{2}m(3)^2 = \frac{1}{2}mv^2 + m(10)(0.25)$

Cancel $m$: $\frac{1}{2}\cdot 9 = \frac{1}{2}v^2 + 2.5$

$4.5 = \frac{1}{2}v^2 + 2.5$

$\frac{1}{2}v^2 = 2$

$v^2 = 4$

$v = 2\ \text{m/s}$

4. Final answer

The speed of the bob at $60^\circ$ to the vertical is: $\boxed{2}$

5. Comparison with stored answer

Stored correct answer = 2

This matches the derived answer.