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

10 m

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20 m

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15 m

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25 m

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Solution

The correct option is A $10 m$

Let maximum height be $H$

From the formula, $v^{2} = u^{2} - 2 g s$
${(10)}^{2} = u^{2} - 2 g H / 2 = u^{2} - g H$ ........ $(1)$

For attaining maximum height, $v = 0$

$0 = u^{2} - 2 g H \Rightarrow u^{2} = 2 g H$

Putting the value of $u^{2}$ in $(1)$ ,

$100 = 2 g H - g H = g H$

$H = \frac{100}{g} = \frac{100}{10} = 10 m e t r e$

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