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' $ .
Then $ v_{v}^{'}=v\sin \theta ={{v}_{x}}=v\cos \theta $ $ v'\frac{\cos \theta }{\sin \theta }=v\cot \theta $ .
Since, $ v_{y}^{'}=gt $ or .
$ t=\frac{{{v}_{y}}-v_{y}^{'}}{g} $ .
$ t=\frac{v\sin \theta -v'\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 } $