Kinematics Question 623
Question: A projectile is thrown in the upward direction making an angle of $ 60{}^\circ $ with the horizontal direction with a velocity of 147$ m{{s}^{-1}} $ . Then the time after Which is inclination with the horizontal it’s $ 45{}^\circ $ , it’s
Options:
A) $ 15( \sqrt{3}-1 )\text{s} $
B)$ 15( \sqrt{3}+1 )\text{s} $
C) $ 7.5( \sqrt{3}-1 )\text{s} $
D)$ 7.5( \sqrt{3}+1 )\text{s} $
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Answer:
Correct Answer: C
Solution:
[c] At the two points of the trajectory during projectile motion, the horizontal component of the velocity is same. Then $ \text{u cos 60 }{}^\circ\text{= v cos 4}5{}^\circ . $
$ \Rightarrow 150\times \frac{1}{2}=\text{v}\times \frac{1}{\sqrt{2}}\text{ or }\frac{150}{\sqrt{2}}\text{ m/s} $
$ \text{Initially: }{{u _{y}}}=u\sin 60{}^\circ =\frac{150\sqrt{3}}{2}\text{ m/s} $
$ \text{Finally: }{{v _{y}}}=v\sin 45{}^\circ =\frac{150}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{150}{2}\text{ m/s} $
$ \text{But }{{v _{y}}}=u _{y}+a _{y}t\text{ or }\frac{150}{2}=\frac{150\sqrt{3}}{2}-10t $
$ 10t=\frac{150}{2}( \sqrt{3}-1 )\text{ or t}\text{= 7}\text{.5}( \sqrt{3}-1 )s $
