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Step-by-Step Solution
Step 1: Identify given quantities
Let the initial velocity of the ball (just as it is thrown upward) be $u$. The ball reaches a maximum height $h$, which implies that
$u = \sqrt{2gh}.$
Here, $g$ is the acceleration due to gravity.
Step 2: Set up the equation of motion
We need to find the times when the ball has covered a vertical distance $h/3$. Consider the motion from the point of projection. Using the equation of motion under constant acceleration,
$ S = ut + \frac{1}{2} a t^2, $
where
$ S = \frac{h}{3},\quad u = \sqrt{2gh}, \quad \text{and} \quad a = -g.
$
Step 3: Substitute the values into the equation
$ \frac{h}{3} = \sqrt{2gh}\,t - \frac{1}{2} g t^2.
$
Rearranging, we get the quadratic equation:
$ \frac{g}{2} t^2 \;-\;\sqrt{2gh}\,t \;+\;\frac{h}{3} = 0.
$
Step 4: Solve the quadratic equation
For a general quadratic equation $A t^2 + B t + C = 0,$ the roots are given by
$ t = \frac{- B \pm \sqrt{B^2 - 4AC}}{2A}.
$
Here,
$ A = \frac{g}{2}, \quad B = -\sqrt{2gh}, \quad C = \frac{h}{3}. $
Substituting and simplifying yields two time values $t_1$ and $t_2$.
Step 5: Determine the ratio of $t_1$ and $t_2$
After simplification, the ratio
$ \frac{t_1}{t_2} $
becomes
$ \frac{\sqrt{2gh} - \sqrt{\frac{4gh}{3}}}{\sqrt{2gh} + \sqrt{\frac{4gh}{3}}}.$
Factor out common terms inside the square roots to obtain:
$ \frac{t_1}{t_2}
= \frac{\sqrt{3}\,\sqrt{\frac{2gh}{3}} - \sqrt{2}\,\sqrt{\frac{2gh}{2}}}
{\sqrt{3}\,\sqrt{\frac{2gh}{3}} + \sqrt{2}\,\sqrt{\frac{2gh}{2}}}
= \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}.
$
Final Answer
$ \displaystyle \frac{t_1}{t_2} = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}.
$
