Q. 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
J & K CETJ & K CET 2009
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 $
