A particle is projected from a point A with a velocity v at an angle of elevation $\theta$ At a certain point B, the particle moves at right angle to its initial direction. Then

$Atinitialpo\mathrm{int},v_x=v\cos \theta and{v}_y=v\sin \theta .$ At second point, where particle moves at right angle to its direction, let its velocity be $v\text{'}.Then,v_x^{\text{'}}=v\sin \theta =v_x=v\cos \theta$

$\Rightarrow v\text{'}\frac{\cos \theta }{\sin \theta }=\cot \theta$

$Since,v_y^{\text{'}}=gt$

$ort=\frac{v_y-v_y^{\text{'}}}{g}$

$\therefore t=\frac{v\sin \theta -v\text{'}\cos \theta }{g}=\frac{v\sin \theta -v\cos \theta .cos\theta }{g}$

$=\frac{v\sin \theta }{g}(1-\cot \theta )=\frac{v}{g\sin \theta }$