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A ball is dropped from a high rise platform at $\mathrm{t}=0$ starting from rest. After $6 \mathrm{~s}$ another ball is thrown downwards from the same platform with a speed $\mathrm{v}$. The two balls meet at $\mathrm{t}=18 \mathrm{~s}$. What is the value of $\mathrm{v}$ ? (take $g=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
The correct answer is:
$74 \mathrm{~ms}^{-2}$
. For first ball, $\mathrm{u}=0$
$$
\therefore \quad \mathrm{s}_1=\frac{1}{2} \mathrm{gt}_1^2=\frac{1}{2} \times \mathrm{g}(18)^2
$$
For second ball, initial velocity $=\mathrm{v}$
$$
\begin{aligned}
& \therefore \mathrm{s}_2=\mathrm{vt}_2+\frac{1}{2} \mathrm{gt}^2 \\
& \Rightarrow \quad \mathrm{t}_2=18-6=12 \mathrm{~s} \\
& \Rightarrow \quad \mathrm{s}_2=\mathrm{v} \times 12+\frac{1}{2} \mathrm{~g}(12)^2
\end{aligned}
$$
Here, $\quad \mathrm{s}_1=\mathrm{s}_2$
$$
\begin{aligned}
\frac{1}{2} g(18)^2 & =12 \mathrm{v}+\frac{1}{2} g(12)^2 \\
\Rightarrow \quad v & =74 \mathrm{~ms}^{-1}
\end{aligned}
$$
$$
\therefore \quad \mathrm{s}_1=\frac{1}{2} \mathrm{gt}_1^2=\frac{1}{2} \times \mathrm{g}(18)^2
$$
For second ball, initial velocity $=\mathrm{v}$
$$
\begin{aligned}
& \therefore \mathrm{s}_2=\mathrm{vt}_2+\frac{1}{2} \mathrm{gt}^2 \\
& \Rightarrow \quad \mathrm{t}_2=18-6=12 \mathrm{~s} \\
& \Rightarrow \quad \mathrm{s}_2=\mathrm{v} \times 12+\frac{1}{2} \mathrm{~g}(12)^2
\end{aligned}
$$
Here, $\quad \mathrm{s}_1=\mathrm{s}_2$
$$
\begin{aligned}
\frac{1}{2} g(18)^2 & =12 \mathrm{v}+\frac{1}{2} g(12)^2 \\
\Rightarrow \quad v & =74 \mathrm{~ms}^{-1}
\end{aligned}
$$
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