question_answer

A particle is projected from a point on the horizontal plane so as to just clear two walls each of height \[20\,m\]at distance \[30\,m\] and \[170\,m\] respectively from the point of projection. If \[\alpha \] is the angle projection then

A)  \[40\,\,\tan \,\alpha =51\]

B)  \[40\,\,\cot \,\alpha =51\]

C)  \[30\,\,tan\,\alpha =23\]

D)  \[30\,\,\cot \,\alpha =23\]

Correct Answer: B

Solution :

Let the angle of projection be \[\alpha \]. Then \[30=u\,\,\cos \,\,\alpha \,t\] \[\Rightarrow \] \[t=\frac{30}{u\,\,\cos \,\alpha }t\] ?.(i) \[\Rightarrow \] \[2\theta =(u\,\sin \,\alpha )t-\frac{1}{2}g{{t}^{2}}\] \[\Rightarrow \]\[20=u\,\,\sin \,\,\alpha \left( \frac{30}{u\,\,\cos \,\alpha } \right)-\frac{1}{2}\,g\,\,{{\left( \frac{30}{u\,\,\cos \,\alpha } \right)}^{2}}\] \[\Rightarrow \] \[20=30\,\tan \alpha -\frac{900}{2}.\frac{g}{{{u}^{2}}\,{{\cos }^{2}}\alpha }\] ?(ii) and \[R=\frac{{{u}^{2}}\,\sin \,2\alpha }{g}=200\] \[\Rightarrow \]\[{{u}^{2}}=\frac{200g}{\sin \,2\,\,\alpha }\] On putting this value in Eq. (ii), we get \[20=30\,\tan \alpha =\frac{900\,g\,\sin \,2\alpha }{2\times 200g\,{{\cos }^{2}}\alpha }\] \[\Rightarrow \] \[20=30\tan \alpha -\frac{9\times 2\,\,\sin \alpha \,\cos \alpha }{4\,{{\cos }^{2}}\alpha }\] \[\Rightarrow \] \[20=30\,\,\tan \,\,\alpha -\frac{9}{2}\,\,\tan \alpha \] \[\Rightarrow \] \[20=\frac{(60-9)}{2}\,\,\tan \,\,\alpha \] \[\Rightarrow \] \[\frac{40}{\tan \alpha }=51\] \[\Rightarrow \] \[40\,\,\cot \,\,\alpha =51\]