A) \[6.7m/sec,81{}^\circ \]To the horizontal
B) \[7.6m/sec,18{}^\circ \]To the horizontal
C) \[5.7m/sec,81{}^\circ \]To the horizontal
D) \[5.7m/sec,18{}^\circ \]
Correct Answer: A
Solution :
Let the initial speed be \[v\,m{{s}^{-1}}.\] |
Using the equation of the path, |
\[y=x\tan \left( -30{}^\circ \right)-\frac{{{x}^{2}}g}{2{{v}^{2}}}{{\sec }^{2}}\left( -30{}^\circ \right)\] |
The stone hits the sea when |
\[\operatorname{y}=-70\,and\,x=20\] |
Therefore,\[-70=\frac{-20}{\sqrt{3}}-\frac{400\times 9.8}{2{{v}^{2}}}\times \frac{4}{3}\] \[\Rightarrow {{v}^{2}}=44.7v=6.7m{{s}^{-1}}\] |
Therefore the initial speed off stone is 6.7m/s. |
To find the direction of motion of the stone we can use |
\[\frac{dy}{dx}=tan\left( -30{}^\circ \right)-\frac{xg\,se{{c}^{2}}\left( -30{}^\circ \right)}{{{v}^{2}}}\] |
When the stone hits the sea, x=20 |
So at that point |
\[\frac{dy}{dx}=\frac{1}{\sqrt{3}}-\frac{20\times 9.8\times 4}{44.7\times 3}=-6.42\]\[\theta ={{\tan }^{-1}}\left( -6.42 \right)\] |
Therefore the stone hits the sea at \[~81{}^\circ \]to the horizontal. |
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