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A student is throwing balls vertically upwards such that he throws the $2^{\text {nd }}$ ball when the $1^{\text {st }}$ ball reaches maximum height. If he throws balls at an interval of 3 second, the maximum height of the balls is $\left(\mathrm{g}=10 \frac{\mathrm{m}}{\mathrm{s}^2}\right)$
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Verified Answer
The correct answer is:
$45 \mathrm{~m}$
At the highest point $\mathrm{v}=0$
$$
\begin{aligned}
& \therefore \mathrm{u}=\mathrm{u}-\mathrm{gt} \\
& \therefore \mathrm{u}=\mathrm{gt}=10 \times 3=30 \mathrm{~ms}
\end{aligned}
$$
Also $0=\mathrm{u}^2-2 \mathrm{gh}$
$$
\therefore \mathrm{h}=\frac{\mathrm{u}^2}{2 \mathrm{~g}}=\frac{(30)^2}{2 \times 10}=45 \mathrm{~m}
$$
$$
\begin{aligned}
& \therefore \mathrm{u}=\mathrm{u}-\mathrm{gt} \\
& \therefore \mathrm{u}=\mathrm{gt}=10 \times 3=30 \mathrm{~ms}
\end{aligned}
$$
Also $0=\mathrm{u}^2-2 \mathrm{gh}$
$$
\therefore \mathrm{h}=\frac{\mathrm{u}^2}{2 \mathrm{~g}}=\frac{(30)^2}{2 \times 10}=45 \mathrm{~m}
$$
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