Physics Two Dimensional Motion Question 161
Question: A projectile is thrown in the upward direction making an angle of $ ,60{}^\circ $ with the horizontal direction with a velocity of $ 147m{{s}^{-1}} $ . Then the time after which its inclination with the horizontal is $ 45{}^\circ $ , is
Options:
A) 15 s B)10.98 s C) 5.49 s D)2.74 s
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Answer:
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, $ {{v _{y}}={{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}( \sqrt{3}-1 )\Rightarrow t=5.49,\text{s} $
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