Question

from a tower of height H, a particle is thrown vertically upwards with speed. 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 1

Dear Student
$$\begin{gathered} Givenheight=H,ballisthrownupwardswithspeeduandtimetakentoreachthel \\ groundisntimesthetimetakentoreachhighestpoint{t}^{\text{'}}=n{t}_1 \\ Lettheheightgainedwhenballisthrownvertically=S \\ v=u+g{t}_1 \\ u=g{t}_1(v=0athighestpoint) \\ {v}^2-u^2=2gS \\ S=\frac{u^2}{2g} \\ whenthevelocityofballiszeroathighestpointthantotaldis\tan cecovere \\ dsay{S}^{\text{'}} \\ {S}^{\text{'}}=Heightoftower+heightgainedbyball \\ {S}^{\text{'}}=H+\frac{u^2}{2g} \\ whentheballisthrownthanfromhighestpointtoground \\ {S}^{\text{'}}=ut+\frac{1}{2}g{t_2}^2 \\ H+\frac{u^2}{2g}=\frac{1}{2}g{t_2}^2(u=0m{s}^{-1}) \\ totaltime=n\times t_1,t^{\text{'}}=n\times t_1 \\ H+\frac{u^2}{2g}=\frac{1}{2}g(t^{\text{'}}-t_1{)}^2 \\ H+\frac{u^2}{2g}=\frac{1}{2}g(n{t}_1-t_1)onsolving \\ {u}^{2}n(n-2)=2gH \\ Regards \end{gathered}$$