The velocity of a projectile when at its greatest height is sqrt((2)/(5)) of its velocity when at half of its greatest height find the angle of projection

The velocity of a projectile when at its greatest height is sqrt((2)/(

1.	The velocity of a projectile when at its greatest height is sqrt((2)/(5)) of its velocity when at half of its greatest height find the angle of projection
Answer» <html><body><p></p>Solution :Step: we know that, <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> of a projectile at half of maximum <a href="https://interviewquestions.tuteehub.com/tag/height-1017806" style="font-weight:bold;" target="_blank" title="Click to know more about HEIGHT">HEIGHT</a> =`usqrt((1+cos^(2)theta)/(2))` <br/> Step 2 : given that `u cos theta=sqrt((2)/(5))xxusqrt((1+cos^(2)theta)/(2))` <br/> <a href="https://interviewquestions.tuteehub.com/tag/squaring-3058178" style="font-weight:bold;" target="_blank" title="Click to know more about SQUARING">SQUARING</a> on both <a href="https://interviewquestions.tuteehub.com/tag/sides-1207029" style="font-weight:bold;" target="_blank" title="Click to know more about SIDES">SIDES</a> `u^(2)cos^(2)theta = (2)/(5)u^(2)((1+cos^(2)theta)/(2))` <br/> `10costheta = 2 + cos theta` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> 8cos^(2)theta = 2 rArr cos^(2)theta=(1)/(4)rArr theta=60^(@)`</body></html>