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 \[\theta =\text{ta}{{\text{n}}^{-\text{1}}}\left(
\text{2} \right)\]
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