From a tower of height H, a particle is thrown verticially upward with a speed u. The time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H, u and n is

From a tower of height H, a particle is thrown verticially upward with

1.	From a tower of height H, a particle is thrown verticially upward with a speed u. The time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H, u and n is
Answer» <html><body><p>2gH=`n^(2)u^(2)`<br/>2gH`=(n-2)^(2)u^(2)`<br/>2gH=`"nu"^(2)(n-2)` <br/>gH`=(n-2)^(2)u^(2)`</p>Solution :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> the heighest point = `(u)/(g)` <br/> Speed on reaching the ground = `sqrt(u^(2)+2gh)` <br/> Now,<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a> = u+at <br/> or, `sqrt(u^(2)+2gh)=-u+"<a href="https://interviewquestions.tuteehub.com/tag/gt-1013864" style="font-weight:bold;" target="_blank" title="Click to know more about GT">GT</a>"` <br/> or, `t= (u+sqrt(u^(2)+2gH))/(g)` <br/> According to the <a href="https://interviewquestions.tuteehub.com/tag/equestion-2618058" style="font-weight:bold;" target="_blank" title="Click to know more about EQUESTION">EQUESTION</a>, <br/> ` (u+sqrt(u^(2)+2gH))/(g)=("nu")/(g)` <br/> or, `2gH=n(n-2)u^(2)` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CHY_DMB_PHY_XI_P1_U02_C01_E07_034_S01.png" width="80%"/></body></html>