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Question

From an elevated point $$P$$, a stone is projected vertically upwards. When the stone reaches a distance $$h$$ below $$P$$, its velocity is double of its velocity at a height $$h$$ above $$P$$. The greatest height attained by the stone from the point of projection $$P$$ is :


A
35h
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B
53h
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C
75h
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D
57h
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E
23h
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Solution

The correct option is B $$\dfrac{5}{3}h$$
From equations of motions, we know
$$v^2=u^2-2gh$$             ...(i)
Here, $$v=2v$$, so
$$(2v)^2=u^2+2gh$$              ...(ii)
Now, on adding Eqs. (i) and (ii), we get
$$5v^2=2u^2$$
$$u^2=\dfrac{5}{2}v^2$$             ...(iii)
On subtracting Eq. (i) in Eq. (ii), we get
$$3v^2=4gh$$
$$v^2=\dfrac{4}{3}gh$$                  ...(iv)
And we know that the
$$H=\dfrac{u^2}{2g}$$
So, here $$H=\dfrac{\dfrac{5}{2}v^2}{2g}=\dfrac{\dfrac{5}{2}\begin{pmatrix}\dfrac{4}{3}gh\end{pmatrix}}{2g}$$
$$H=\dfrac{5}{3}h$$

Physics

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