A bob of a simple pendulum has a speed of 3 m/s at its lowest...

Let the lowest position be the reference level for potential energy. At the lowest position, the total mechanical energy is purely kinetic:

$E_i = \dfrac{1}{2} m v_0^2$

When the pendulum makes an angle $\theta$ with the vertical, the height of the bob above the lowest position is:

$h = L - L \cos\theta = L(1 - \cos\theta)$

At this position, the total mechanical energy is the sum of kinetic and potential energy:

$E_f = \dfrac{1}{2} m v^2 + mgh$

By conservation of mechanical energy:

$\dfrac{1}{2} m v_0^2 = \dfrac{1}{2} m v^2 + mgh$

$v^2 = v_0^2 - 2gh$

Substituting $h = L(1 - \cos\theta)$:

$v^2 = v_0^2 - 2gL(1 - \cos\theta)$

Given $v_0 = 3$ m/s, $L = 0.5$ m, \theta = 60^{}^{\circ}, and $g = 10$ m/s$^2$:

v^2 = (3)^2 - 2 \times 10 \times 0.5 \times (1 - \cos 60^{}^{\circ})

$v^2 = 9 - 10 \times \left(1 - \dfrac{1}{2}\right)<br><br>$ v^2 = 9 - 10 $\times \dfrac{1}{2}<br><br>$ v^2 = 9 - 5 = 4$<br><br>$v = 2$m/s<br><br>Answer: 2$ m/s