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» 2gH=`n^(2)u^(2)` 2gH`=(n-2)^(2)u^(2)` 2gH=`"nu"^(2)(n-2)` gH`=(n-2)^(2)u^(2)` Solution :Time taken to REACH the heighest point = `(u)/(g)` Speed on reaching the ground = `sqrt(u^(2)+2gh)` Now,V = u+at or, `sqrt(u^(2)+2gh)=-u+"GT"` or, `t= (u+sqrt(u^(2)+2gH))/(g)` According to the EQUESTION, ` (u+sqrt(u^(2)+2gH))/(g)=("nu")/(g)` or, `2gH=n(n-2)u^(2)`