A) \[\frac{u\sin \alpha }{f}\]
B) \[\frac{f\cos \alpha }{u}\]
C) \[u\sin \alpha \]
D) \[\frac{u\cos \alpha }{f}\]
Correct Answer: D
Solution :
We have, \[{{R}^{2}}={{u}^{2}}+{{f}^{2}}{{t}^{2}}+2uft\cos \,({{180}^{o}}-\alpha )\] \[{{R}^{2}}={{u}^{2}}+{{f}^{2}}{{t}^{2}}-2uft\cos -\alpha \] Let \[V={{u}^{2}}+{{f}^{2}}{{t}^{2}}-2uft\cos \alpha \] \[\frac{dV}{dt}=0=2{{t}^{2}}t-2utf\cos \alpha \] \[\frac{{{d}^{2}}V}{d{{t}^{2}}}=2{{t}^{2}}=+ve\] i.e., velocity will be least after a time. \[\frac{dV}{dt}=0=2{{f}^{2}}t-2uf\cos \alpha \] \[\Rightarrow 2{{f}^{2}}t=2uf\cos \alpha \] \[\Rightarrow \] \[t=\frac{2uf\cos \alpha }{2{{f}^{2}}}\] \[\Rightarrow \] \[t=\frac{u\cos \alpha }{f}\]You need to login to perform this action.
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