question_answer

A particle starts from origin with speed u and angle \[\alpha \] with horizontal. Its velocity is perpendicular to initial velocity after time

A) \[\frac{u}{g}\sin \alpha \]                       

B) \[\frac{u}{g}\sec \alpha \]

C) \[\frac{u}{g}\cos \alpha \]                      

D) \[\frac{u}{g}\text{cosec}\,\alpha \]

Correct Answer: D

Solution :

Let particle is moving perpendicular to its initial direction after time t.

After time t velocity is let v, then velocity components of particle at time t are

\[{{v}_{{{x}_{f}}}}=v\sin (90-\alpha )=v\cos \alpha \]

As there is no change in magnitude of horizontal component, \[{{v}_{{{x}_{f}}}}={{v}_{{{x}_{i}}}}\]

\[v\cos \alpha =ucos\alpha \]

And vertical component at instant t is

\[-\,v\sin (90-\alpha )=u\sin \alpha -gt\]

From these equations, we have

\[-\,cot\alpha =\frac{u\sin \alpha -gt}{u\cos \alpha }\]

\[\Rightarrow \]\[\frac{gt}{u\cos \alpha }=\frac{1}{\cos \alpha sin\alpha }\]

\[\Rightarrow \]\[t=\frac{u}{g\sin \alpha }\]or \[t=\frac{u}{g}\text{cosec}\alpha \]