Question

please solve this one..

Answer 1

hope it helps u......

Answer 2

Hope u like my process
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In a projectile, the velocity of the particle consists of 2 components i.e. Horizontal and vertical component. and can be divided in
$= > v \cos( \beta ) \: \: and \: \: v \sin( \beta )$
respectively.

Now,.

At top most height, the velocity of vertical component becomes 0 .. But horizontal component remains the same..

So

=> Kinetic Energy at top most height

$= \frac{1}{2} m {(v \: \cos( \beta ) )}^{2}$

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At initial point the velocity at both components are same... So,.. Velocity remains same.

Thus, kinetic energy at initial position

$= \frac{1}{2} m {v}^{2}$


Now.. Given that

=>Energy at top = ¾ of energy at initial.

$= > \frac{1}{2} m {(v \cos( \beta )) }^{2} = \frac{3}{4} \times \frac{1}{2} m {v}^{2} \\ \\ = > \cos {}^{2} ( \beta ) = \frac{3}{4} \\ \\ = > \cos( \beta ) = \sqrt{ \frac{3}{4} } = \frac{ \sqrt{3} }{2} \\ \\ = > thus \: \\ \\ = > \beta = \bf \underline{30}$
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Thus the value of beta is 30°
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