question_answer

A particle just clears a wall of height b at a distance a and strikes the ground at a distance c from the point of projection. The angle of projection is       AIEEE  Solved  Paper-2007

A)  \[{{\Lambda }^{o}}\]           

B)         \[{{\Lambda }^{o}}\]                    

C)         \[NaCl\]

D)         \[{{\Lambda }^{o}}\]

Correct Answer: C

Solution :

\[a=(u\cos \alpha )t\]and\[b=(u\sin \alpha )t-\frac{1}{2}g{{t}^{2}}\] \[b=a\tan \alpha -\frac{1}{2}g\frac{{{a}^{2}}}{{{u}^{2}}{{\cos }^{2}}\alpha }\] Also,    \[c=\frac{{{u}^{2}}\sin 2\alpha }{g}\] \[b=a\tan \alpha -\frac{{{a}^{2}}g}{2}\left( \frac{\sin 2\alpha }{cg} \right){{\sec }^{2}}\alpha \] \[\Rightarrow \]\[b=a\tan \alpha -\frac{{{a}^{2}}}{2c}\times 2\tan \alpha \] \[\Rightarrow \]\[\left( a-\frac{{{a}^{2}}}{c} \right)\tan \alpha =b\] \[\Rightarrow \]\[\tan \alpha =\frac{bc}{a(c-a)}\] \[\Rightarrow \]\[\alpha ={{\tan }^{-1}}\frac{bc}{a(c-a)}\]