A body of $1 kg$ is thrown vertically upwards with an initial velocity $(u) = 2 m/s$ , the height at which kinetic energy of the body is $(\frac{1}{4})^{th}$ of its original value is (take $g = 10 {m/s}^{2}$ )

0.15 m

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0.25 m

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0.3 m

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0.35 m

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Solution

The correct option is A $0.15 m$
Given, initial velocity $(u) = 2 m/s$
Mass of body $(m) = 1 kg$
$∴$ Initial kinetic energy $(K_{i}) = \frac{1}{2} m u^{2} = \frac{1}{2} \times 1 \times (2)^{2} = 2 J$
Final kinetic energy $(K_{f}) = \frac{1}{4}$ Initial kinetic energy $(K_{i})$
From the law of conservation of energy,
$U_{i} + K_{i} = U_{f} + K_{f}$
$U_{i} = 0 J; U_{f} = m g h J$
$∴ K_{i} = U_{f} + K_{f}$
$U_{f} = K_{i} - K_{f}$
$= K_{i} - \frac{K_{i}}{4} = \frac{3 K_{i}}{4} = \frac{3}{4} \times 2 = \frac{3}{2} J$
$m g h = \frac{3}{2} J$
$\Rightarrow h = \frac{3}{2 \times m g} = \frac{3}{2 \times 1 \times 10} = \frac{3}{20} = 0.15 m$

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A body of 1 kg is thrown vertically upwards with an initial velocity (u)=2 m/s, the height at which kinetic energy of the body is (14)th of its original value is (take g=10 m/s2)

A body of $1 kg$ is thrown vertically upwards with an initial velocity $(u) = 2 m/s$ , the height at which kinetic energy of the body is $(\frac{1}{4})^{th}$ of its original value is (take $g = 10 {m/s}^{2}$ )