question_answer

A stone is thrown from the top of a cliff 70m high at an angle of \[30{}^\circ \] below the horizontal and hits the sea 20m from the bottom of the cliff Find the initial speed of the stone and the direction in which it is moving when it hits the sea.

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.