Question

From a tower of height $H$, a particle is thrown vertically 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

• A. $2gH=n{u}^2(n-2)$
• B. $gH=(n-2)u^2$
• C. $2gH=n^2{u}^2$
• D. $gH=(n-2)u^2$

Answer 1

The correct option is A $2gH=n{u}^2(n-2)$

$t_1:$ time taken to reach the highest point
$t_1=\frac{u}{g}...\left(1\right)$
$t$ : time taken to reach the ground
Applying kinematics equation,
$-H=ut-\frac{1}{2}g{t}^2$
$\Rightarrow t^2-\frac{2u}{g}t-\frac{2H}{g}=0$
$t=\frac{\frac{2u}{g}\pm \sqrt{\frac{4u^2}{g^2}+\frac{8H}{g}}}{2}$
$t=\frac{u}{g}\pm \sqrt{\frac{u^2}{g^2}+\frac{2H}{g}}$
$\Rightarrow t=t_1+\sqrt{t_1^2+\frac{2H}{g}}$(from $\left(1\right)$)
As $t>t_1,$ $-ve$ sign is neglected.
Given $t=n{t}_1$
$n{t}_1=t_1+t_1\sqrt{1+\frac{2H}{g{t}_1^2}}$
$n=1+\sqrt{1+\frac{2gH}{u^2}}$
$(n-1{)}^2=1+\frac{2gH}{u^2}$
$\left[\right(n-1{)}^2-1]=\frac{2gH}{u^2}$
$\Rightarrow 2gH=\left[\right(n-1{)}^2-1]u^2$
$\Rightarrow 2gH=\left[n\right(n-2\left)\right]u^2$