A bucket of water, of mass 2.5 kg in all, is whirled on a rope in a vertical circle of radius 1.2 m…

Step 1. The top. With $T = 0$: $mg = \dfrac{mv_t^2}{r}$, so $v_t = \sqrt{rg} = \sqrt{11.76\ \text{m}^2/\text{s}^2} = 3.43\ \text{m/s}$.

Step 2. The bottom speed. $v_b^2 = 11.76\ \text{m}^2/\text{s}^2 + 4 \times 9.8\ \text{m/s}^2 \times 1.2\ \text{m} = 58.8\ \text{m}^2/\text{s}^2$, so $v_b = 7.67\ \text{m/s}$.

Step 3. Tension at the bottom. $T = m\left(g + \dfrac{v_b^2}{r}\right) = 2.5\ \text{kg} \times (9.8\ \text{m/s}^2 + 49\ \text{m/s}^2) = 147\ \text{N}$.