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Q. When a particle is thrown vertically upwards, its velocity at one third of its maximum height is $10\, \sqrt{2} m / s$. The maximum height attained by it is

Motion in a Straight Line

Solution:

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$v^{2}-u^{2}=-2\, g \times \frac{2 H}{3}$
$\Rightarrow -100 \times 2=-2 \times 10 \times \frac{2 H}{3}$
$\Rightarrow H=15\, m$