Question

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 (1) velocity of particle at B is $v\sin \theta$ (2) velocity of particle at B is $v\cot \theta$ (3) velocity of particle at B is $v\tan \theta$ (4) velocity of flight from A to B is $\frac{v}{g\sin \theta }$

• A. 1, 2 and 3 are correct
• B. 1 and 2 are correct
• C. 2 and 4 are correct
• D. 1 and 3 are correct

Answer 1

Answer: (C)

At initial point $v_x=v\cos \theta$ and $v_y=\sin \theta$ . At second point, where particle moves at right angle to its direction let its velocity be $v^{\prime }$ .
Then $v_v^{{}^{\prime }}=v\sin \theta =v_x=v\cos \theta$ $v^{\prime }\frac{\cos \theta }{\sin \theta }=v\cot \theta$ .
Since, $v_y^{{}^{\prime }}=gt$ or .
$t=\frac{v_y-v_y^{{}^{\prime }}}{g}$ .
$t=\frac{v\sin \theta -v^{\prime }\cos \theta }{g}$
$=\frac{v\sin \theta -v\cot \theta \cos \theta }{g}$
$=\frac{v\sin \theta }{g}(1-\cot \theta )=\frac{v}{g\sin \theta }$