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 :
Average velocity\[=\frac{Displacement}{Time}\] \[{{v}_{av}}=\sqrt{\frac{{{H}^{2}}+\frac{{{R}^{2}}}{4}}{\frac{T}{2}}}\] Here, \[H=\]maximum height\[=\frac{{{v}^{2}}{{\sin }^{2}}\theta }{2g}\] \[R=\]range\[=\frac{{{v}^{2}}\sin 2\theta }{g}\] and \[T=\]time of flight\[=\frac{2v\sin \theta }{g}\] \[\therefore \] \[{{v}_{av}}=\frac{v}{2}\sqrt{1+3{{\cos }^{2}}\theta }\]You need to login to perform this action.
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