Question

please solve this problem.

Answer 1

refer to the attachment

Answer 2

Hope u like my process
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We know,

$= > \bf \: horizontal \: \: range = \frac{ {u}^{2} \sin(2 \theta ) }{g} \\ \\ = > \bf \: greatest \: \: height \: = \frac{ {(u \sin( \theta ) )}^{2} }{2g}$

Given,
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=> Horizontal range = 2 × greatest height.

So,

$= > \it \frac{ {u}^{2} \sin(2 \theta ) }{g} = 2 \times \frac{ {u}^{2} \sin {}^{2} ( \theta ) }{2g} \\ \\ = > \it \: \frac{ {u}^{2} \times 2 \sin( \theta ) \cos( \theta ) }{g} = \frac{ {u}^{2} \times 2 \sin {}^{2} ( \theta ) }{2g} \\ \\ = > \bf \: \sin(2 \theta ) = { \sin }^{2} ( \theta )......(1)\\ \\ = > \cos( \theta ) = \frac{ \sin( \theta ) }{2} \\ \\ = > \sqrt{1 - { \sin}^{2}( \theta) } = \frac{ \sin( \theta ) }{2} \\ \\ = > 1 - \sin {}^{2} ( \theta ) = \frac{1}{4} \sin {}^{2} ( \theta ) \\ \\ = > \frac{5}{4} \sin {}^{2} ( \theta ) = 1 \\ \\ = > \sin( \theta ) = \sqrt{ \frac{4}{5} } = \bf \underline{ \frac{2}{ \sqrt{5} } }$

Hence the range of the projectile

$= > \bf \: \frac{ {v}^{2} \sin(2 \theta ) }{g} \\ \\ = \it \frac{ {v}^{2} \sin {}^{2} ( \theta ) }{g}...... (by \:\: 1)\\ \\ = \frac{ {v}^{2} \times {( \frac{2}{ \sqrt{5} } )}^{2} }{g} \\ \\ = > \bf \underline{ \frac{4 {v}^{2} }{5g} }......( \it \: answer)$

Hence option a (✔️) is the correct answer.
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Hope this is ur required answer

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