Question

answer please anyone ​

Answer 1

Answer:

Let the angle of projection be alpha and initial velocity be u.

Horizontal velocity is fix and its

$= u \cos( \alpha )$

Maximum height

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

When the height half of it, vertical velocity

$2as = {v}^{2} - {u}^{2} \\ \\ - 2g( \frac{ {u}^{2} \sin {}^{2} ( \alpha ) }{4g} ) = {v}^{2} - {u}^{2} \sin {}^{2} ( \alpha ) \\ \\ v = \sqrt{ \frac{1}{2} } u \sin( \alpha )$

Which is given

$\frac{1}{ \sqrt{2} } u \sin( \alpha ) \times \sqrt{ \frac{2}{5} } = u \cos( \alpha ) \\ \\ \tan( \alpha ) = \sqrt{5} \\ \\ \alpha = \tan {}^{ - 1} ( \sqrt{5} )$

Note

If answer is given 60° then there is misprint and there is 3 instead of 5 in question