A) \[\frac{v}{2}\sqrt{1+2{{\cos }^{2}}\theta }\]
B) \[\frac{v}{2}\sqrt{1+2{{\cos }^{2}}\theta }\]
C) \[\frac{v}{2}\sqrt{1+3{{\cos }^{2}}\theta }\]
D) \[v\cos \theta \]
Correct Answer: C
Solution :
Initial velocity \[=v\cos \theta \hat{i}+v\sin \theta \hat{j}\] Final velocity \[=v\cos \theta \,\hat{i}\] The\[x\]component of the velocity remains \[v\cos \theta ,\]so its average value is also \[v\cos \theta ,\]whereas the y component of the velocity changes from \[v\sin \theta \]to, zero, so its average value is\[\frac{v\sin \theta }{2}.\] \[\therefore \]Average velocity\[=v\cos \theta \,\hat{i}+\frac{v\sin \theta \hat{j}}{2}\] Magnitude of average velocity \[=\frac{1}{2}\sqrt{4{{v}^{2}}{{\cos }^{2}}\theta +{{v}^{2}}{{\sin }^{2}}\theta }\] \[=\frac{1}{2}\sqrt{{{v}^{2}}+3{{v}^{2}}{{\cos }^{2}}\theta }=\frac{v}{2}\sqrt{1+3{{\cos }^{2}}\theta }\] Hence, the correction option is [c].
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