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Key ideas you must know

1. Vertical motion with constant acceleration

   • Equation of motion: $y = ut - \tfrac12 g t^2$ where $u$ is the launch speed, $g$ is gravitational acceleration, $y$ is height after time $t$.

2. Maximum height

   • At the highest point the velocity becomes zero:
     $0 = u - g t_{top} \;\Rightarrow\; t_{top} = \frac{u}{g}$
   • The corresponding height is
     $H = \frac{u^2}{2g}$

3. Two times for the same height

   • Because the height equation is quadratic in $t$, a specific height below $H$ is reached twice (ascending & descending). The two roots of the quadratic give those instants.

Logical chain to solve

1. Express the required height, say $y = H/3$.
2. Substitute $u^2 = 2gH$ into $y = ut - \tfrac12 g t^2$.
3. Divide by $H$ to get a dimension-less quadratic in a new variable $x = \dfrac{g t}{u}$.
4. Solve the quadratic; you automatically get two positive roots $x_1$ (up) and $x_2$ (down).
5. Because $t \propto x$, the ratio of times is the same as the ratio of $x$ values.
6. Simplify to an elegant surd form for the final ratio.

What’s the story?

Imagine you throw a ball straight up. It goes up, slows down, stops for a split-second at the top, then falls back down.

Now draw an imaginary balcony at one-third of the top height. The ball passes that balcony twice:

1. On the way up
2. On the way down

Because gravity keeps pulling the ball the whole time, those two instants don’t happen at equal intervals from the throw. The question simply asks:

When does the ball cross that balcony while going up and while coming down, and what is the ratio of those two times?

Answer in one line

The ratio (up : down) turns out to be
$\displaystyle (3-\sqrt{6}) : (3+\sqrt{6}) \;\approx\; 1 : 9.9$

Step-by-step solution

1. Symbols
   Initial speed $u$, gravitational acceleration $g$, maximum height $H$. $H = \frac{u^2}{2g} \quad\Rightarrow\quad u = \sqrt{2gH}$

2. Height at one-third of the top
   $y = \frac{H}{3}$

3. Equation of motion
   $\frac{H}{3} = u t - \frac12 g t^2$

4. Insert $u$ and divide by $H$
   $\frac13 = \frac{ \sqrt{2gH}\, t}{H} - \frac12 \frac{g t^2}{H}$ Using $H = \dfrac{u^2}{2g} \;\Rightarrow\; \dfrac{1}{H} = \dfrac{2g}{u^2}$, $\frac13 = \frac{2g t}{u} - \frac{g^2 t^2}{u^2}$

5. Make it dimension-less
   Let $x = \frac{g t}{u}$ (so $t = \dfrac{u}{g} x$). Then $\frac13 = 2x - x^2$ i.e. $x^2 - 2x + \frac13 = 0$

6. Solve the quadratic
   $x = \frac{2 \pm \sqrt{4 - \tfrac43}}{2} = 1 \pm \sqrt{\frac23}$ Both roots are positive: $x_1 = 1 - \sqrt{\frac23}, \qquad x_2 = 1 + \sqrt{\frac23}$

7. Times for the two events
   $t_1 = \frac{u}{g} x_1 \quad(\text{ascending}), \qquad t_2 = \frac{u}{g} x_2 \quad(\text{descending})$

8. Required ratio
   Because $u/g$ cancels, $\boxed{\dfrac{t_1}{t_2} = \dfrac{x_1}{x_2} = \frac{1 - \sqrt{\tfrac23}}{1 + \sqrt{\tfrac23}}}$ Put the surd over a common denominator ($3$): $x_1 = \frac{3 - \sqrt6}{3}, \quad x_2 = \frac{3 + \sqrt6}{3}$ So the neat integer-free form is $\boxed{(3 - \sqrt6) : (3 + \sqrt6)} \;\approx\; 1 : 9.9$