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A ball is dropped from the top of the building $100 \mathrm{~m}$ high. Simultaneously, another ball is thrown upwards from the bottom of the building with such a velocity that the balls collide exactly mid-way. What is the speed in $\mathrm{ms}^{-1}$ with which the second ball is thrown? (Take, $g=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
The correct answer is:
$31.6$
The given situation is shown in the figure.
For the first ball, time taken to reach at mid point $A$,
$$
\begin{aligned}
t &=\sqrt{\frac{2 h}{g}} \\
&=\sqrt{\frac{2 \times 50}{10}}=\sqrt{10} \mathrm{~s}
\end{aligned}
$$
For the second ball,
$$
\begin{aligned}
v &=0, u=?, a=-g \\
\therefore \text { From equation, } v &=u+g t \\
0 &=u-g t_{1} \\
\Rightarrow \quad t_{1} &=\frac{u}{g}
\end{aligned}
$$
Since, both balls collide at some time,
Hence,
$$
\begin{aligned}
t &=t_{1} \\
\sqrt{10} &=\frac{u}{g} \\
u &=10 \sqrt{10} \mathrm{~m} / \mathrm{s}=31.6 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
$$
\Rightarrow
$$
For the first ball, time taken to reach at mid point $A$,
$$
\begin{aligned}
t &=\sqrt{\frac{2 h}{g}} \\
&=\sqrt{\frac{2 \times 50}{10}}=\sqrt{10} \mathrm{~s}
\end{aligned}
$$
For the second ball,
$$
\begin{aligned}
v &=0, u=?, a=-g \\
\therefore \text { From equation, } v &=u+g t \\
0 &=u-g t_{1} \\
\Rightarrow \quad t_{1} &=\frac{u}{g}
\end{aligned}
$$
Since, both balls collide at some time,
Hence,
$$
\begin{aligned}
t &=t_{1} \\
\sqrt{10} &=\frac{u}{g} \\
u &=10 \sqrt{10} \mathrm{~m} / \mathrm{s}=31.6 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
$$
\Rightarrow
$$
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