1. **Problem statement:** A man throws balls vertically upwards with the same speed at intervals of 2 seconds. We want to find the minimum speed so that more than two balls are in the sky at any time. 2. **Relevant formula:** The time a ball stays in the air when thrown vertically upwards with speed $v$ is given by $$t = \frac{2v}{g}$$ where $g = 9.8$ m/s² is the acceleration due to gravity. 3. **Explanation:** - Each ball stays in the air for time $t$. - Balls are thrown every 2 seconds. - To have more than two balls in the air simultaneously, the time $t$ must be greater than $2 \times 2 = 4$ seconds because if $t \leq 4$, at most two balls can be in the air. 4. **Calculate minimum speed:** $$t > 4$$ $$\frac{2v}{9.8} > 4$$ Multiply both sides by 9.8: $$2v > 39.2$$ Divide both sides by 2: $$v > 19.6$$ 5. **Interpretation:** The speed must be greater than 19.6 m/s to have more than two balls in the sky at any time. **Final answer:** The speed of the throw should be more than 19.6 m/s.