JEE Main · 2007 · 29 Apr 2007 · Offline · Chapter_pending

From projectile motion the path is $y=x\tan \alpha -\frac{g{x}^2}{2u^2{\cos }^2\alpha }$ and the range is $c=\frac{u^2\sin 2\alpha }{g}$. Clearing the wall means $b=a\tan \alpha -\frac{g{a}^2}{2u^2{\cos }^2\alpha }$, and using the range to replace $\frac{u^2}{g}$ turns this into $b=\left(a-\frac{a^2}{c}\right)\tan \alpha$. Solving for the angle gives $\tan \alpha =\frac{bc}{a(c-a)}$, so the projection angle is ${\tan }^{-1}\frac{bc}{a(c-a)}$.