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A particle is released from a height $H$. At a certain height, its kinetic energy is half of its potential energy with reference to the surface of the earth. Height and speed of the particle at that instant are respectively
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Verified Answer
The correct answer is:
$\frac{2 H}{3}, \sqrt{\frac{2 g H}{3}}$
Total mechanical energy of particle, i.e.
Given,
$$
\mathrm{KE}=\frac{1}{2} \mathrm{PE}
$$
i.e.
$$
\begin{aligned}
& \frac{\mathrm{KE}}{\mathrm{PE}}=\frac{1}{2} \\
& \mathrm{PE}=2 \mathrm{KE}
\end{aligned}
$$
or
Substituting in this Eq. (i), we get
$$
\begin{aligned}
\mathrm{PE}+\mathrm{KE} & =m g H \\
2 \mathrm{KE}+\mathrm{KE} & =m g H \\
3 \mathrm{KE} & =m g H \\
\mathrm{KE} & =\frac{m g H}{3} \\
\mathrm{PE} & =\frac{2}{3} m g H
\end{aligned}
$$
Similarly,
So, height from ground at that instant,
$$
h=\frac{2 H}{3}
$$
So, speed of particle,
$$
v=\sqrt{2 g h}=\sqrt{2 g H / 3}
$$
Given,
$$
\mathrm{KE}=\frac{1}{2} \mathrm{PE}
$$
i.e.
$$
\begin{aligned}
& \frac{\mathrm{KE}}{\mathrm{PE}}=\frac{1}{2} \\
& \mathrm{PE}=2 \mathrm{KE}
\end{aligned}
$$
or
Substituting in this Eq. (i), we get
$$
\begin{aligned}
\mathrm{PE}+\mathrm{KE} & =m g H \\
2 \mathrm{KE}+\mathrm{KE} & =m g H \\
3 \mathrm{KE} & =m g H \\
\mathrm{KE} & =\frac{m g H}{3} \\
\mathrm{PE} & =\frac{2}{3} m g H
\end{aligned}
$$
Similarly,
So, height from ground at that instant,
$$
h=\frac{2 H}{3}
$$
So, speed of particle,
$$
v=\sqrt{2 g h}=\sqrt{2 g H / 3}
$$
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