Q. A stone is projected vertically upward to reach maximum height $h$ . The ratio of its kinetic energy to its potential energy at a height $\frac{4}{5}h$ , will be
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Solution:
At a height $\frac{4 h}{5}$ , the potential energy of the stone will be,
$PE=mg\left(\frac{4 h}{5}\right)...\left(1\right)$
The total energy of the stone will be $mgh$ , hence the kinetic energy of the particle will be,
$KE=mgh-\frac{4}{5}mgh=\frac{m g h}{5}...\left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$ ,
$\frac{K E}{P E}=\frac{\frac{m g h}{5}}{\frac{4 m g h}{5}}=\frac{1}{4}$ .
$PE=mg\left(\frac{4 h}{5}\right)...\left(1\right)$
The total energy of the stone will be $mgh$ , hence the kinetic energy of the particle will be,
$KE=mgh-\frac{4}{5}mgh=\frac{m g h}{5}...\left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$ ,
$\frac{K E}{P E}=\frac{\frac{m g h}{5}}{\frac{4 m g h}{5}}=\frac{1}{4}$ .