Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
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
Correct Answer: C
Solution :
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 \[t=\frac{{{v}_{y}}-v_{y}^{'}}{g}\] \[t=\frac{v\sin \theta -v'\cos \theta }{g}\] \[=\frac{v\sin \theta -v\cot \theta \cos \theta }{g}\] \[=\frac{v\sin \theta }{g}(1-\cot \theta )=\frac{v}{g\,\sin \theta }\]
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