Question

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.

• A. $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
• B. $\frac{\sqrt{3}-1}{\sqrt{3}+1}$
• C. $\frac{1}{3}$
• D. $\frac{\sqrt{2}-1}{\sqrt{2}+1}$

Hint:

For a projectile motion under gravity, the time equation $s = ut - \frac{1}{2}gt^2$ is a quadratic. The two roots represent the times to pass a certain height, once on the way up and once on the way down. The ratio is required here, not the individual times.

Answer 1

The Correct Option is A

Let the initial velocity be u. At the maximum height h, the final velocity is 0.
Using the equation of motion $v^2 = u^2 + 2as$:
$0 = u^2 - 2gh \implies u = \sqrt{2gh}$.
Now, consider the time t to reach a height $s = h/3$. Using $s = ut + \frac{1}{2}at^2$:
$\frac{h}{3} = (\sqrt{2gh})t - \frac{1}{2}gt^2$.
This is a quadratic equation for time t:
$\frac{1}{2}gt^2 - (\sqrt{2gh})t + \frac{h}{3} = 0$.
The two roots of this equation, $t_1$ and $t_2$, represent the times when the ball is at height $h/3$ (once going up, and once coming down).
Using the quadratic formula, $t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$t = \frac{\sqrt{2gh} \pm \sqrt{(\sqrt{2gh})^2 - 4(\frac{g}{2})(\frac{h}{3})}}{2(\frac{g}{2})}$.
$t = \frac{\sqrt{2gh} \pm \sqrt{2gh - \frac{2gh}{3}}}{g}$.
$t = \frac{\sqrt{2gh} \pm \sqrt{\frac{4gh}{3}}}{g}$.
$t = \frac{\sqrt{g}(\sqrt{2h} \pm \sqrt{4h/3})}{g} = \frac{\sqrt{h}}{g}(\sqrt{2g} \pm \frac{2\sqrt{g}}{\sqrt{3}})$.
Let's factor out $\sqrt{2gh}$ from the numerator:
$t = \frac{\sqrt{2gh}}{g} \left( 1 \pm \sqrt{\frac{4gh/3}{2gh}} \right) = \sqrt{\frac{2h}{g}} \left( 1 \pm \sqrt{\frac{2}{3}} \right)$.
The two times are $t_1 = \sqrt{\frac{2h}{g}} \left( 1 - \sqrt{\frac{2}{3}} \right)$ and $t_2 = \sqrt{\frac{2h}{g}} \left( 1 + \sqrt{\frac{2}{3}} \right)$.
The ratio of these times is:
$\frac{t_1}{t_2} = \frac{1 - \sqrt{2/3}}{1 + \sqrt{2/3}} = \frac{1 - \frac{\sqrt{2}}{\sqrt{3}}}{1 + \frac{\sqrt{2}}{\sqrt{3}}}$.
$\frac{t_1}{t_2} = \frac{\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}}{\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}}} = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$.