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 \[t=\frac{u}{g}\text{cosec}\alpha \] |
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