An aeroplane is rising vertically with acceleration a. Two stones are dropped from it at an interval of time t. The distance between them at time t' after the second stone is dropped will be

\frac{1}{2} t [g t + 2 g t^{'} + 2 a t^{'}]

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\frac{1}{2} (g + a) (t + 2 t^{'}) t

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\frac{1}{2} (g + a) (t - t^{'})^{2}

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\frac{1}{2} (g + a) (t + t^{'})^{2}

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Solution

The correct option is A $\frac{1}{2} t [g t + 2 g t^{'} + 2 a t^{'}]$
$\begin{matrix} t = t^{'} h e i g h t o f B f r o m g r o u n d = v (t + t^{'}) - \frac{1}{2} g {(t + t^{'})}^{2} h e i g h t o f A f r o m g r o u n d = (v + a t) t^{'} - \frac{1}{2} g t^{' 2} ∴ h = {v (t + t^{'}) - \frac{1}{2} g {(t + t^{'})}^{2}} - {(v + a t) t^{'} - \frac{1}{2} g t^{' 2}} = ∣ ∣ v t + v t^{'} - \frac{1}{2} g t^{2} - \frac{1}{2} g t^{' 2} - g t t^{'} - v t^{'} - a t t^{'} + \frac{1}{2} g t^{' 2} ∣ ∣ = ∣ ∣ v t - \frac{1}{2} g t^{2} - g t t^{'} - a t t^{'} ∣ ∣ N o w, t a k i n g i n i t i a l v e l o c i t y o f p l a n e a s z e r o h = \frac{1}{2} g t^{2} + g t t^{'} + a t t^{'} = \frac{1}{2} t [g t + 2 g t^{'} + 2 a t^{'}] \end{matrix}$
Hence,
Option A is correct answer.

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