Question

please solve 34 and 36.

Answer 1

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Formula to be used :
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$= > \bf r = \frac{ {u}^{2} \sin(2 \theta ) }{g} \\ \\ where \: \: \: \: r = range \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: u = velocity \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \theta = angle \: \: of \: \: projection \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: g = gravity$
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34)

=> Range for 1st projection

$= \frac{ {u}^{2} \sin2( 45 - \theta )}{g } = \frac{ {u}^{2} \sin(90 - 2 \theta ) }{g} \\ \\ = \bf \: \frac{ {u}^{2} \cos(2 \theta ) }{g}$

=> Range for 2nd projection.

$= \frac{ {u}^{2} \sin2(45 + \theta ) }{g} = \frac{ {u}^{2} \sin(90 + 2 \theta ) }{g} \\ \\ = \bf \: \frac{ {u}^{2} \cos(2 \theta ) }{g}$

So, the ratio of the ranges of projection 1st to projection 2nd is

$= \bf \: \frac{ {u}^{2} \cos( 2\theta ) }{g} : \frac{ {u}^{2} \cos(2 \theta ) }{g} = \underline{1 :1}$
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36)

Smaller the $\sin2{\theta}$ shortest the range would be..

So,

For P,

$= > \sin(2 \theta ) = \sin(2 \times 15) \\ \\ = \sin(30) = \bf \underline{\frac{1}{2}}$

For Q,

$= > \sin(2 \theta ) = \sin(2 \times 30) \\ \\ = \sin(60) = \bf \underline{ \frac{ \sqrt{3} }{2} }$
For R,

$= > \sin(2 \theta ) = \sin(2 \times 45) \\ \\ = \sin(90) = \bf \underline{1}$

For S,

$= > \sin(2 \theta ) = \sin(2 \times 60) \\ \\ = \sin(120) = \bf \underline{\frac{ \sqrt{3} }{2}}$
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Thus the value is least at 15°

Hence the body has shortest horizontal range at an angle 15°

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