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A particle is thrown vertically upwards. If its velocity at half of the maximum height is $10 \mathrm{~m} / \mathrm{s}$, then maximum height attained by it is (Take $\left.g=10 \mathrm{~m} / \mathrm{s}^2\right)$
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10 m
Let particle thrown with velocity $u$ and its maximum height is $H$ then $H=\frac{u^2}{2 g}$ When particle is at a height $H / 2$, then its speed is $10 \mathrm{~m} / \mathrm{s}$
From equation $v^2=u^2-2 g h$
$\begin{aligned} & (10)^2=u^2-2 g\left(\frac{H}{2}\right)=u^2-2 g \frac{u^2}{4 g} \Rightarrow u^2=200 \\ & \text { Maximum height } \Rightarrow H=\frac{u^2}{2 g}=\frac{200}{2 \times 10}=10 \mathrm{~m} \\ & \end{aligned}$
From equation $v^2=u^2-2 g h$
$\begin{aligned} & (10)^2=u^2-2 g\left(\frac{H}{2}\right)=u^2-2 g \frac{u^2}{4 g} \Rightarrow u^2=200 \\ & \text { Maximum height } \Rightarrow H=\frac{u^2}{2 g}=\frac{200}{2 \times 10}=10 \mathrm{~m} \\ & \end{aligned}$
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