question_answer

A projectile is thrown in the upward direction making an angle of \[\,60{}^\circ \] with the horizontal direction with a velocity of \[147\text{ }m{{s}^{-1}}\]. Then the time after which its inclination with the horizontal is \[45{}^\circ \], is

A) 15 s                 

B)        10.98 s

C) 5.49 s               

D)        2.74 s

Correct Answer: C

Solution :

[c] Velocity of projectile \[\text{u = 147 m}{{\text{s}}^{-1}}\]

Angle of projection \[\alpha =60{}^\circ \]

Let, the time taken by the projectile from O to A be t where direction \[\beta =45{}^\circ \]. As horizontal component of velocity remains constant during the projectile motion.

\[\Rightarrow \text{v}\,\,\cos 45{}^\circ =\text{u cos}\,\text{60}{}^\circ \]

\[\Rightarrow \text{v}\times \frac{1}{\sqrt{2}}=147\times \frac{1}{2}\Rightarrow \text{v}=\frac{147}{\sqrt{2}}\text{m}{{\text{s}}^{-1}}\]

For Vertical motion, \[{{\text{v}}_{y}}={{\text{u}}_{y}}-\text{gt}\]

\[\Rightarrow \,\,\text{v}\sin 45{}^\circ =45\sin \,60{}^\circ -9.8\,\text{t}\]

\[\Rightarrow \frac{147}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=147\times \frac{\sqrt{3}}{2}-9.8\,\text{t}\]\[\Rightarrow \text{9}\text{.8}\,\text{t}\,\text{=}\,\,\frac{147}{2}\left( \sqrt{3}-1 \right)\Rightarrow t=5.49\,\text{s}\]