A body projected vertically from ground with initial velocity u. At half of its maximum height h0 (in its upward journey) its acceleration reduces from g downwards to g/2 downwards, then the value of u2h0 is

V2=u2+2gh2 ... (i) V2=2gh4 ...... (ii)⇒u2=3gh2⇒u2h=3g2

A body projected vertically from ground with initial velocity u. At half of its maximum height h0 (in its upward journey) its acceleration reduces from g downwards to g/2 downwards, then the value of u2h0 is

V2=u2+2gh2 ... (i) V2=2gh4 ...... (ii)⇒u2=3gh2⇒u2h=3g2

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A body projected vertically from ground with initial velocity u. At half of its maximum height h₀ (in its upward journey) its acceleration reduces from g downwards to g/2 downwards, then the value of $\frac{u^{2}}{h_{0}}$ is

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g/2

3g/2

answer is C.

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$V^{2} = u^{2} + \frac{2 g h}{2} . . . (i) \begin{array}{l} V^{2} = \frac{2 g h}{4} . . . . . . (i i) \\ \Rightarrow u^{2} = \frac{3 g h}{2} \\ \Rightarrow \frac{u^{2}}{h} = \frac{3 g}{2} \end{array}$

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A body projected vertically from ground with initial velocity u. At half of its maximum height h0 (in its upward journey) its acceleration reduces from g downwards to g/2 downwards, then the value of u2h0 is

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A body projected vertically from ground with initial velocity u. At half of its maximum height h₀ (in its upward journey) its acceleration reduces from g downwards to g/2 downwards, then the value of $\frac{u^{2}}{h_{0}}$ is