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 the time taken by it to reach the highest point of its path. The relation between $H$, $u$ and $n$ is

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

Answer 1

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

Let $t_1$ and $t_2$ be the time taken by the particle to reach maximum height and ground respectively.
Since final velocity will be zero when particle reaches maximum height, therefore time taken to reach maximum height, $t_1=\frac{u}{g}$


Using second equation of motion, we can say
$\begin{array}{}-H=u{t}_2-\frac{1}{2}g{t}_2^2& \text{(1)}\end{array}$
$\begin{array}{}{t}_2=n\times t_1& \text{(2)}\end{array}$
Using $\left(2\right)$ in $\left(1\right)$, we get
$-H=u\frac{nu}{g}-\frac{1}{2}g\frac{n^2{u}^2}{g^2}$
$⟹-H=\frac{2n{u}^2-n^2{u}^2}{2g}$
$\therefore 2gH=n{u}^2(n-2)$