Animated Solution for Physics - Kinematics: 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 Kinematics Journey

Imagine you are standing at the edge of a tall tower of height $H$. You hold a small stone in your hand and toss it straight up into the air with an initial velocity $u$. It climbs, slows down, stops for a brief magical moment, and then plunges all the way down past you, finally striking the ground below.

This classic physics scenario holds a beautiful mathematical secret, and our goal is to uncover the relationship between the tower's height $H$, your throwing speed $u$, and a special time ratio $n$.

The Power of the Origin

Before we write a single equation, we must establish our frame of reference. This is where many students make a fatal error. Let's place our origin $(y=0)$ exactly at the point of projection—the top of the tower. We will define the upward direction as positive $(+y)$.

This simple choice makes our math incredibly clean. It means that any position above the tower is positive, and any position below the tower is negative. Consequently, when the stone finally hits the ground, its net displacement is not $H$, but $-H$.

Reaching the Peak

Let's analyze the first part of the journey. The stone travels upwards until gravity saps all its kinetic energy. At the maximum height, its velocity $v$ becomes exactly zero.

Let's call the time taken to reach this peak $t_1$. Using the first equation of motion, $v=u+at$, we can easily find this time:

This is a fundamental result. The time to reach the top depends only on how fast you threw it and the pull of gravity.

Connecting the Dots

Now, let's look at the entire flight from the moment it leaves your hand to the moment it smashes into the ground. Let's call this total time $t_2$.

The problem gives us a fascinating constraint: the total time $t_2$ is $n$ times the time taken to reach the peak $t_1$. Mathematically, this means:

To link this time to the height of the tower, we use the workhorse of kinematics—the second equation of motion:

We must substitute our values into this equation with absolute precision regarding signs. The net displacement $S$ is $-H$, the initial velocity is $+u$, the acceleration $a$ is $-g$, and the time $t$ is our total time $t_2$.

Now, we bring in our expression for $t_2$:

Let's carefully expand the squared term. Notice how one $g$ in the numerator will cancel with one $g$ in the denominator:

This equation looks a bit messy with those denominators and negative signs. Let's clean it up by multiplying the entire equation by $-2g$. This elegant move clears the fractions and flips the signs in one go:

Finally, we factor out the common term $n{u}^2$ on the right side to reveal the hidden relationship:

And there it is! A pristine, beautiful equation connecting the height of the tower, the initial velocity, and the time ratio. This perfectly matches option (c). Notice how this equation also tells us a physical truth: for the height $H$ to be positive, the ratio $n$ must be strictly greater than 2. Physics and math, working in perfect harmony!