Question

From a tower of height $H$, a particle is thrown vertically upwards 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. $n^2{u}^2=2gH$
• B. $(n-2{)}^2{u}^2=gH$
• C. $n(n-2)u^2=2gH$
• D. $(n-2)u^2=gH$

Answer 1

The correct option is C $n(n-2)u^2=2gH$

The given situation can be shown as


Speed of the particle on reaching the ground, $v=\sqrt{u^2+2gH}$
Also, $v=u+at\Rightarrow \sqrt{u^2+2gH}=-u+gt$
(taking downward direction positive)
Or, $t=\frac{u+\sqrt{u^2+2gH}}{g}$......(i)

We know that, time taken to reach the highest point is $t=\frac{u}{g}$.....(ii)
From (i) and (ii) we have
$\frac{u+\sqrt{u^2+2gH}}{g}=n\times \frac{u}{g}$
(from question)
$\Rightarrow u(n-1)=\sqrt{u^2+2gH}$
Squaring both sides, we get
$u^2(n-1{)}^2=u^2+2gH$
$\Rightarrow n(n-2)u^2=2gH$