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, \[{{\text{v}}_{y}}={{\text{u}}_{y}}-\text{gt}\] |
\[\Rightarrow \,\,\text{v}\sin 45{}^\circ =45\sin \,60{}^\circ -9.8\,\text{t}\] |
\[\Rightarrow \frac{147}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=147\times \frac{\sqrt{3}}{2}-9.8\,\text{t}\]\[\Rightarrow \text{9}\text{.8}\,\text{t}\,\text{=}\,\,\frac{147}{2}\left( \sqrt{3}-1 \right)\Rightarrow t=5.49\,\text{s}\] |
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