A) \[4\cos \theta \]
B) \[4(2\cos \theta -\sec \theta )\]
C) \[2(\sec \theta +4\cos \theta )\]
D) \[4(\sec \theta +\cos \theta )\]
Correct Answer: B
Solution :
Key Idea: Horizontal component of velocity remains same. Let \[v\] be the velocity, when projected with angle\[\theta \], then equating the horizontal velocities in both the cases, we get \[v\cos \theta =u\cos 2\,\,\theta \] \[\Rightarrow \] \[v=\frac{u\cos 2\theta }{\cos \theta }\] where \[\sec \theta =\frac{1}{\cos \theta }\] \[\therefore \] \[v=u\cos 2\theta \sec \theta \] Using \[\cos 2\theta =2{{\cos }^{2}}\theta -1,\] we get Given, \[u=4\,\,m/s,\] we get \[v=4(2{{\cos }^{2}}\theta -1)\sec \theta \] \[\Rightarrow \] \[v=4(2\cos \theta -\sec \theta )\]You need to login to perform this action.
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