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 points of its path. What is the relation between $H$, $u$, and $n$?

Answer 1

Step 1: Given

Height of the tower is $H$.
Let $t_1$ be the time taken to reach maximum height and $t_2$ be the time taken to fall to the ground from maximum height. Then, as per given information,
$t_2=n{t}_1$

Step 2: Formulas used

We know that, the time taken for a particle to reach maximum height is,
$t_m=\frac{u\sin \theta }{g}$
where $u$ is the initial velocity and $g$is the acceleration due to gravity.

From the three equations of motion, the distance covered is given as,
$s=ut+\frac{1}{2}a{t}^2$ where $a$ is the acceleration of the body

Step 3: Apply the distance formula from three laws of motion

We know that the particle is thrown directly upwards, thus the time taken to reach maximum height is
$t_1=\frac{u\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{g}=\frac{u}{g}$

In the distance formula,
$s=-H,t=t_2,a=-g$
Substituting the values in the formula for distance covered, we get

$\begin{array}{ccc}\Rightarrow & -H=un{t}_1-\frac{1}{2}g{\left(n{t}_1\right)}^2& \left[\because t_2=n{t}_1\right]\\ \Rightarrow & -H=un\frac{u}{g}-\frac{1}{2}g{n}^2{\left(\frac{u}{g}\right)}^2& \left[\because t_1=\frac{u}{g}\right]\\ \Rightarrow & 2gH=-2u^{2}n+n^2{u}^2& \\ \therefore & 2gH=n{u}^2\left(n-2\right)& \end{array}$

Therefore, the relationship between $H$, $u$, and $n$ is $2gH=n{u}^2\left(n-2\right)$.