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
B
53h
C
75h
D
57h
E
23h

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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