question_answer

The angle of projection for which the horizontal range is equal to the maximum height of a projectile is :

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,         \[R=H,\] \[\therefore \]  \[\frac{{{u}^{2}}\sin 2\theta }{g}=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] \[\Rightarrow \]               \[2\sin \theta \cos \theta =\frac{{{\sin }^{2}}\theta }{2}\] \[\Rightarrow \]               \[4\cos \theta =\sin \theta \] \[\Rightarrow \]               \[\tan \theta =4\] \[\Rightarrow \]               \[\theta ={{\tan }^{-1}}(4)\]