A man throws balls with the same speed vertically upwards one after the other at an interval of $2 ext{ seco

Question

A man throws balls with the same speed vertically upwards one after the other at an interval of $2 	ext{ seconds}$. What should be the speed of the throw so that more than two balls are in the sky at any time? (Given $g=9.8 	ext{ m/s}^2$)

Accepted Answer

1. **Analyze the Condition:** The problem states that balls are thrown at intervals of $\Delta t = 2 	ext{ s}$. For 'more than two balls' (i.e., at least 3) to be in the sky simultaneously, the first ball must remain in the air until the third ball is thrown.
2. **Time Constraint:** Let the time of flight for a single ball be $T$. If the third ball is thrown at $t = 2\Delta t = 4 	ext{ s}$, the first ball (thrown at $t=0$) must not have landed yet. Therefore, the time of flight must be greater than $4 	ext{ s}$ ($T > 4 	ext{ s}$).
3. **Formula for Time of Flight:** For an object thrown vertically upwards with speed $u$ and returning to the same level, the total time of flight is given by $T = \frac{2u}{g}$ .
4. **Calculation:**
   $$ \frac{2u}{g} > 4 $$
   $$ 2u > 4g $$
   $$ u > 2g $$
   Substitute $g = 9.8 	ext{ m/s}^2$:
   $$ u > 2 	imes 9.8 $$
   $$ u > 19.6 	ext{ m/s} $$
5. **Conclusion:** The speed of the throw must be more than $19.6 	ext{ m/s}$.

Step-by-Step Solution

Practice All PHYSICS Questions