A stone is thrown vertically upwards. When the particle is at one-half its maximum height, its speed is $10 m / s$ , then maximum height attained by particle is $(g = 10 m / s^{2})$ :

8 m

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10 m

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15 m

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20 m

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Solution

The correct option is B $10 m$
From third equation of motion
$v^{2} = u^{2} - 2 g h$
Given v = 10m/s at h/2. But v=0, when particle attains maximum height h.
Therefore, $10^{2} = u^{2} - 2 g (\frac{h}{2})$
Now, $u^{2} - 0 = 2 g h o r u^{2} = 2 g h$
$\Rightarrow 100 = 2 g h - g h = g h$
$\Rightarrow h = 10 m$

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