Kinematics Question 613
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 is inclination with the horizontal it’s $ 45{}^\circ $ , it’s
Options:
A) 15 s
B)10.98 s
C) 5.49 s
D)2.74 s
Show Answer
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} $
BETA
