A) \[\theta ={{\tan }^{-1}}(1)\]
B) \[\theta ={{\tan }^{-1}}\](4)
C) \[\theta ={{\tan }^{-1}}\](3)
D) \[\theta ={{\tan }^{-1}}\](2)
Correct Answer: B
Solution :
Let a particle be projected with initial velocity u at an angle\[\theta \]with the horizontal. Then, range (R) of projectile is \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] ?.. (i) Maximum height \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] ?. (ii) Given, \[R=H,\] \[\therefore \] \[\frac{{{u}^{2}}\sin 2\theta }{g}=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] \[\Rightarrow \] \[2\sin \theta \cos \theta =\frac{{{\sin }^{2}}\theta }{2}\] \[\Rightarrow \] \[4\cos \theta =\sin \theta \] \[\Rightarrow \] \[\tan \theta =4\] \[\Rightarrow \] \[\theta ={{\tan }^{-1}}(4)\]You need to login to perform this action.
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