question_answer

One second after the projection, a stone moves at an angle of \[\text{45}{}^\circ \] with the horizontal. Two seconds from the start, it is travelling horizontally. Find the angle of projection with the horizontal.

Answer:

                After one second, let \[{{\upsilon }_{x}}\], \[{{\upsilon }_{y}}\]be the horizontal and vertical component velocities of the projectile whose initial velocity of projection u and angle of projection is \[\theta \], then \[{{\upsilon }_{\text{x}}}=\text{u}\,\text{cos}\theta \]and \[{{\upsilon }_{y}}=\text{u}\,\text{sin}\,\theta -\text{ g}\times \text{1}=u\,\sin \theta -g\] As the resultant of \[{{\overset{\to }{\mathop{\upsilon }}\,}_{x}}\] and \[{{\overset{\to }{\mathop{\upsilon }}\,}_{y}}\]akes an angle  \[\beta \left( =\text{45}{}^\circ  \right)\] with the horizontal, so \[\frac{{{\upsilon }_{y}}}{{{\upsilon }_{x}}}=\frac{u\sin \theta -g}{u\cos \theta }=\tan {{45}^{o}}\]or \[u\,\sin \theta -2g=0\] or \[\text{u}(\text{sin}\theta \text{cos}\theta )=\text{g}\]                                         .......(i) After two seconds, the vertical component velocity of projectile becomes zero, since the velocity of projectile is horizontal after two seconds. So \[\text{u sin}\theta \text{2g}=0\]or \[\text{u}=\text{ 2g}/\text{sin}\theta \] From (i), \[\frac{2g}{\sin \theta }(\text{sin}\theta \text{cos}\theta )=\text{gor 2 }(\text{1 }-\text{ cotq})\text{ }=\text{1}\] or            \[\text{1}-\text{cot}\theta =\frac{1}{2}\] or            \[\text{cot}\theta =\text{1}\frac{1}{2}=\frac{1}{2}\] or            \[\text{tan}\theta =\text{2}\] or            \[\theta =\text{ta}{{\text{n}}^{-\text{1}}}\left( \text{2} \right)\]