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A stone is projected from level ground with speed u and ann at angle theta with horizontal. Somehow the acceleration due to gravity (g) becomes double (that is 2g) immediately after the stone reaches the maximum height and remains same thereafter. Assume direction of acceleration due to gravity always vertically downwards. Q. The horizontal range of particle is

Answer» <html><body><p>`(3)/(4)(<a href="https://interviewquestions.tuteehub.com/tag/u-1435036" style="font-weight:bold;" target="_blank" title="Click to know more about U">U</a>^(2)sin2theta)/(g)` <br/>`(u^(2)sin 2theta)/(2g) (1+(1)/(<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(2)))`<br/>`(u^(2))/(g) sin 2theta` <br/>`(u^(2)sin2theta)/(2g) (2+(1)/(sqrt(2)))`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :The time taken to <a href="https://interviewquestions.tuteehub.com/tag/reach-1178062" style="font-weight:bold;" target="_blank" title="Click to know more about REACH">REACH</a> 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> and maximum height are `t=(usintheta)/(g)andH=(u^(2)sin^(2)theta)/(2g)` <br/> For remaining half, the time of flight is <br/> `t'=sqrt((2H)/((2g)))=sqrt((u^(2)sin^(2)theta)/(2g^(2)))=(t)/(sqrt(2))` <br/> `:.` Total time of flight is `t+t',T=t(1+(1)/(sqrt(2)))` <br/> `T=(usintheta)/(g)=(1+(1)/(sqrt(2)))` <br/> So horizontal range is u `costhetaxxT=(u^(2)sin2theta)/(2g)(1+(1)/(sqrt(2)))`</body></html>


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