A) \[{{\sin }^{-1}}\left( \frac{4}{5} \right)\]
B) \[{{\sin }^{-1}}\left( \frac{3}{5} \right)\]
C) \[{{\sin }^{-1}}\left( \frac{4}{3} \right)\]
D) \[{{\sin }^{-1}}\left( \frac{3}{4} \right)\]
Correct Answer: A
Solution :
\[x=36t\]\ \[{{v}_{x}}=\frac{dx}{dt}=36\,m/s\] \[y=48t-4.9{{t}^{2}}\]\ \[{{v}_{y}}=48-9.8t\] at \[t=0\] \[{{v}_{x}}=36\] and \[{{v}_{y}}=48\,m/s\] So, angle of projection \[\theta ={{\tan }^{-1}}\left( \frac{{{v}_{y}}}{{{v}_{x}}} \right)={{\tan }^{-1}}\left( \frac{4}{3} \right)\] Or \[\theta ={{\sin }^{-1}}(4/5)\]
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