Kinematics Question 637
Question: The position of a projectile launched from the origin at $ t=0 $ it’s given by $ \vec{r}=(40\hat{i}+50\hat{j})m $ at 2s. If the projectile was launched at an angle $ \theta $ from the horizontal, then $ \theta $ it’s (take $ \text{g = 10 m}{{\text{s}}^{-2}} $ )
Options:
A) $ {{\tan }^{-1}}\frac{2}{3} $
B)$ {{\tan }^{-1}}\frac{3}{2} $
C) $ {{\tan }^{-1}}\frac{7}{4} $
D)$ {{\tan }^{-1}}\frac{4}{5} $
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Answer:
Correct Answer: C
Solution:
[c] From question,
Horizontal velocity (initial), $ {{\text{u}} _{\text{x}}}\text{=}\frac{\text{40}}{\text{2}}\text{=20 m/s} $
Vertical velocity (initial),$ \text{50=}{{\text{u}} _{\text{y}}}\text{t+}\frac{\text{1}}{\text{2}}\text{g}{{\text{t}}^{\text{2}}} $
$ \Rightarrow {{\text{u}} _{\text{y}}}\times 2+\frac{1}{2}( -10 )\times 4;\text{ or, }{{u _{y}}}=\frac{70}{2}=35\text{ m/s;} $
$ \therefore \tan \theta =\frac{{{\text{u}} _{\text{y}}}}{{{\text{u}} _{\text{x}}}}=\frac{35}{20}=\frac{7}{4}\Rightarrow \text{ Angle}\theta \text{=}\text{ta}{{\text{n}}^{-1}}\frac{7}{4} $
