A particle is thrown vertically upwards. Its velocity at half of the height is $10$ $m / s$ . The maximum height attained by the particle is $(g = 10 m / s^{2})$ :

8 m

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20 m

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10 m

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16 m

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Solution

The correct option is C $10 m$
$v^{2} = u^{2} - 2 g h$
for maximum height $v = 0$ and $h = H$
so $2 g H = u^{2}$
again at half of the height
$(10)^{2} = u^{2} - 2 g (H / 2) \Rightarrow 100 = 2 g H - g H$
$∴ g H = 100 \Rightarrow H = \frac{100}{10} = 10 m$

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