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 point of projection. The initial velocity will be

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$\frac{gt}{2 \sin (α - β)}$

$\frac{gtcos β}{2 \sin (α - β)}$

$\frac{\sin (α - β)}{2 gt}$

$\frac{2 \sin (α - β)}{gtcos β}$

answer is B.

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Detailed Solution

See figure (2).

Let u be the velocity of projection. Now

$\begin{array}{l} y = (usin α) t - \frac{1}{2} {gt}^{2} \\ x = (ucos α) t \end{array}$

From figure,

$\begin{array}{l} \tan β = \frac{y}{x} \\ \tan β = \frac{(usin α) t - \frac{1}{2} {gt}^{2}}{(ucos α) t} \\ = \tan α - \frac{gt}{2 ucos α} \end{array}$

Solving, we get

$\begin{array}{l} u = \frac{gt}{2 \cos α (\tan α - \tan β)} \\ = \frac{gtcos αcos β}{2 \cos α [\sin αcos β - \sin βcos α]} \\ = \frac{gtcos β}{2 \sin (α - β)} \end{array}$