Question

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

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Solution

**Step 1: Given**

Time interval: $t=2s$

Initial velocity is $u$.

Final velocity: $v=0$ (It will come to rest in the end).

Acceleration is $a$.

Acceleration due to gravity: $g=9.8m/{s}^{2}$.

**Step 2: Formula used**

According to the first equation of motion the final velocity is $v=u+at$, where $u$ is initial velocity, $a$ is acceleration and $t$ is time. Here the ball is thrown upwards so it is going against gravity, thus the acceleration is due to gravity, so the formula can be written as $v=u-gt$ (negative because it is acting against gravity by moving upwards).

**Step 3: Solution**

Replace the acceleration, $a$, in the equation with acceleration due to gravity, $-g$ (negative because it is acting against gravity by moving upwards).

$v=u+\left(-g\right)t\phantom{\rule{0ex}{0ex}}=u-gt$

Substitute$v=0$, $g=9.8m/{s}^{2}$ and $t=2s$ in the equation $v=u-gt$ to find the velocity:

$0=u-9.8\times 2\phantom{\rule{0ex}{0ex}}\Rightarrow 0=u-19.6\phantom{\rule{0ex}{0ex}}\Rightarrow u=19.6m/{s}^{2}$

This is the velocity required to have 2 balls in the air at the same time. If more than 2 balls should be in the air, then the velocity should be more than this. Because at a velocity just greater than this, the first ball will be just above the ground before landing, the second ball will be at the top point and the third ball will be just thrown.

**Hence, the velocity should be more than **$19.6m/{s}^{2}$**, so that more than two balls are in the sky at any time.**

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