At the highest point of the vertical circle, the net force directed towards the center is the sum of the tension ($T$) in the string and the gravitational force ($mg$), as both forces point downwards. This net force provides the centripetal force, $F_c = \frac{mv^2}{r}$, required to keep the ball on its circular path.

Applying Newton's second law in the radial direction:

$$ \sum F_r = T + mg = \frac{mv^2}{r} $$

where $r$ is the length of the string, and $v$ is the speed of the ball.

[Q1] To find the tension as a multiple of the ball's weight ($W=mg$), we first derive an expression for $T$.

$$ T = \frac{mv^2}{r} - mg $$

Now, we express this as a multiple of weight by dividing by $mg$:

$$ \frac{T}{mg} = \frac{v^2}{rg} - 1 $$

Substituting the given values, $v = 5$ m/s, $r = 1$ m, and using $g = 9.8$ m/s²:

$$ \frac{T}{mg} = \frac{(5 \text{ m/s})^2}{(1 \text{ m})(9.8 \text{ m/s}^2)} - 1 = \frac{25}{9.8} - 1 \approx 2.551 - 1 = 1.551 $$

So, the tension is approximately 1.55 times the ball's weight.

[Q2] The minimum speed ($v_{min}$) to stay on the circular path occurs when the tension in the string becomes zero ($T=0$). If the speed were any lower, the string would go slack. We use the same force equation and set $T=0$:

$$ 0 + mg = \frac{mv_{min}^2}{r} $$

Solving for $v_{min}$:

$$ v_{min}^2 = gr $$ $$ v_{min} = \sqrt{gr} $$

Substituting the values $r = 1$ m and $g = 9.8$ m/s²:

$$ v_{min} = \sqrt{(9.8 \text{ m/s}^2)(1 \text{ m})} = \sqrt{9.8} \text{ m/s} \approx 3.13 \text{ m/s} $$