Animated Solution for Physics - Kinematics: A ball is thrown up with a certain velocity, so that it reaches a height $h$. Find the ratio of the two different times of the ball reaching $\frac{h}{3}$ in both the directions.

Imagine throwing a ball straight up into the air. It leaves your hand, fights against gravity, slows down until it momentarily stops at its maximum height $h$, and then accelerates back down to the ground. If we plot its height $y$ against time $t$, we get a beautiful, symmetric downward-facing parabola.

Now, pick a specific height, say $\frac{h}{3}$. If you draw a horizontal line across your parabolic graph at $y=\frac{h}{3}$, you'll see it intersects the curve at exactly two points. These two points correspond to the two times the ball is at that height: $t_1$ during its upward journey, and $t_2$ during its downward journey. Our goal is to find the ratio $\frac{t_1}{t_2}$.

Before we can find the times, we need to know the initial velocity $u$ with which the ball was thrown. We know that at the maximum height $h$, the final velocity $v$ is zero. Using the third equation of motion:

Substituting $v=0$ and $a=-g$, we get:

Now we have our initial velocity in terms of the maximum height. To connect height and time, we bring in the second equation of motion:

We want to find the times when the displacement $s$ is $\frac{h}{3}$. Substituting $s=\frac{h}{3}$, $u=\sqrt{2gh}$, and $a=-g$, we get:

Let's rearrange this equation to make it look like a standard quadratic equation $a{t}^2+bt+c=0$:

This is the mathematical heart of the problem. The fact that it's a quadratic equation perfectly mirrors the physical reality that there are two times for a given height! We can solve for $t$ using the quadratic formula:

Plugging in our coefficients:

The two roots represent our two times, $t_1$ (the smaller time, using the minus sign) and $t_2$ (the larger time, using the plus sign). We need their ratio:

The $g$ in the denominator cancels out immediately. Even better, we can factor out $\sqrt{2gh}$ from both the numerator and the denominator:

To make this look like the options provided, we multiply the numerator and denominator by $\sqrt{3}$:

And there we have it! A beautifully elegant result that depends only on the fraction of the maximum height, completely independent of the actual height or the strength of gravity.