Physics, asked by ritik12336, 11 months ago

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Answers

Answered by Anonymous
55
\huge\underline\mathfrak{Answer}

1:1

\textbf{Step-By-Step-Solution}

\textbf{Formula used}

Horizontal Range

r = \frac{ {u}^{2} \sin(2) \alpha }{g}

\textbf{Given}

\textbf{Angle of projection }

 \alpha (1) = 45 - \alpha

 \alpha (2) = 45 + \alpha

\textbf{Using , Formula}

\textbf{Step 1}

For first motion the eq become

r = \frac{ {u}^{2} \sin(2 ) \alpha (1)}{g}

 \alpha (1) = 45 - \alpha

\textbf{Eq 1}

r = \frac{ {u}^{2} \sin2(45 - \alpha ) }{g}

\textbf{Step 2}

For second motion the eq become

r = \frac{ {u}^{2} \sin(2 ) \alpha (2)}{g}

\alpha (1) = 45 + \alpha

\textbf{Eq 2}

r = \frac{ {u}^{2} \sin2(45 + \alpha ) }{g}

\textbf{Step 3}

\textbf{Ratio}

Divide eq (1) by eq(2)

after cancelled the value of \textbf{g}

 \frac{r(1)}{ r(2) } = \frac{ {u}^{2} \sin2(45 - \alpha ) }{ {u}^{2} \sin2(45 + \alpha ) }

 \frac{r(1)}{ r(2) } = \frac{ \sin2(45 - \alpha ) }{ \sin2(45 + \alpha ) }

 \frac{r(1)}{ r(2) } = \frac{ (\sin90 - \alpha ) }{ \sin(90 + \alpha ) }

 \frac{r(1)}{r(2)} = \frac{cos2 (\alpha) }{ \cos2( \alpha ) }

\textbf{Hence}

<marquee>

Horizontal range ratio= 1:1
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