A particle is projected at an angle of elevation $α$ , and after t second, it appears to have an angle of elevation $β$ as seen from the point of projection. The initial velocity will be

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$\frac{\sin (α - β)}{2 g t}$

$\frac{g t \cos β}{2 \sin (α - β)}$

$\frac{g t}{2 \sin (α - β)}$

$\frac{2 \sin (α - β)}{g t \cos β}$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{y}{x} = \tan β \\ \Rightarrow \frac{usin αt - \frac{1}{2} {gt}^{2}}{ucos αt} = \tan β \end{array}$

$or usin α - \tan βucos α = \frac{gt}{2} or u = \frac{gtcos β}{2 \sin (α - β)}$

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