Physics, asked by sumanshakya255, 5 months ago

Man is running with the speed of 10m/s and finds the rain is falling vertically with speed of 20m/s. At what speed should the man run, such that the rain appear to him, at an angle of 37 degree
with vertical.

Answers

Answered by MrAnonymous412
11

 \\  \underline  \orange{\sf \: Requiréd  \: Question} :  -  \\

➣ Man is running with the speed of 10m/s and finds the rain is falling vertically with speed of 20m/s. At what speed should the man run, such that the rain appear to him, at an angle of 37° with vertical.

 \\  \underline  \orange{\sf \: Requiréd  \: Answer} :  -  \\

 \:  \:  \:  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \sf\longrightarrow \:  \:  \:  \bold{ \frac{5}{2} } \sqrt{3}

 \\  \underline{ \large  \green{\sf \: Given \: } :  - } \\

  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: ➬  \:  \:  \:  \: \sf The \: speed\: of\:  man \: V_m \: = 10  \: m/s  \\

  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: ➬  \:  \:  \:  \: \sf The \: speed\: of\:  the \: rain \:  \: V_r \: = 20  \: m/s  \\

  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: ➬  \:  \:  \:  \: \sf angle \:  = 37 \degree  \\

 \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \: ➬  \:  \:  \:  \: \sf V_r ^{ \rightarrow } sin \theta \:  = V_m^{ \rightarrow }  \\ \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \sf 20 \times  sin \theta \:  = 10 \\  \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:   \underline{\sf\theta \:  = sin ^{ - 1}  \frac{1}{2}  = 30 \degree} \\

 \large\underline  \green{\sf \: solution \: } :  -  \\

Now,

 \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \sf \: V_m^{ \rightarrow } \:  = V^{ \rightarrow } \\

 \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    \:  :  \implies\:  \:  \sf \: V_rcos30° = R^{-1}cos37° \\

Where are is resultant ,

 \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \sf \:  20 \times  \frac{ \sqrt{3} }{2} \:  \:  =  \:  \:   R \times  \frac{4}{5} \\

 \\  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \boxed{\pink{\frak{\:     R  =  \frac{25 \sqrt{3} }{2}} }} \\

Now ,

 \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    \:   \color{grey}{\:  \:  \sf \: V_rcos30° = R^{-1}cos37°} \\

 \\   \:  \:  \:  \:  \rightarrow \:  \:  \sf \: 20 \times   \: \frac{1}{2}   = \frac{25 \sqrt{3} }{2}  \times  \frac{3}{5} \\

 \\   \:  \:  \:  \:  \rightarrow  \bold{ \: \:  \: V_m    = 10  \:  -  \frac{15}{2}   \sqrt{3} }\\

 \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \pink{\boxed{ \rightarrow  \bold{ \: \:  \: V_m    =   \frac{5}{2}  \sqrt{3} }}}\\

 \sf \frac{5}{2}  \sqrt{3}\\ m/s should the man run .

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