A ball is thrown vertically upwards. It was observed at a height h twice with a time interval $Δ t$ . The initial velocity of the ball is

\sqrt{8 g h + (g Δ t)^{2}}

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\sqrt{2 g h + {(\frac{g Δ t}{2})}^{2}}

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\sqrt{8 g h + (2 g Δ t)^{2}}

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\sqrt{2 g h}

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Solution

The correct option is D $\sqrt{2 g h + {(\frac{g Δ t}{2})}^{2}}$
$Let the ball be at height h at time 't', then,$
$h = u t - \frac{1}{2} g t^{2}$
$\Rightarrow t^{2} - \frac{2 u}{g} t + \frac{2 h}{g} = 0$
$This has two roots t_{1} & t_{2}$
$\Rightarrow t_{1} t_{2} = \frac{2 h}{g} & t_{1} + t_{2} = \frac{2 u}{g}$
$Let t_{2} - t_{1} = Δ t$
$So, (Δ t)^{2} = (t_{1} + t_{2})^{2} - 4 t_{1} t_{2} = {(\frac{2 u}{g})}^{2} - 4 (\frac{2 h}{g})$
$\Rightarrow {(\frac{2 u}{g})}^{2} = (Δ t)^{2} + 4 (\frac{2 h}{g})$
$\Rightarrow u = \sqrt{2 g h + {(\frac{g Δ t}{2})}^{2}}$

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