How long will it take a stone to fall from the top of a tower? A stone dropped from the top of a tower is found to travel $5/9$ of the height of the tower during the last second of the fall. The time of fall is?
Let $T$ be the time it takes for the stone to reach the ground and $h$ be the height of the tower.
$$h=5T^2.$$
Let the distance covered in $T-1$ seconds be $x$.
$$x=5(T-1)^2,$$
as $g=10$.
Since $h-x$ is the distance covered in the last second$=5/9$
$$\frac 59 = 5(T^2-T^2-1+2t) \\
\frac 19 = 2T-1 \\
18T=10 \\
t= \frac 59$$
And it’s the wrong answer. I can’t seem to point what’s going wrong, so I could really use some help with that.
 A: The height covered the last second isn't $\frac 59\approx 0.56$. It's 5/9 of the height of the tower, i.e. $\frac59h$.
Using that, we get
$$
\frac59h = \frac{25}9T^2 = 5(T^2-T^2+2T-1)
$$
which may be solved like a regular quadratic equation. It has two solutions, as many quadratic equations do. One may be discarded because it says the ball has travelled less than a second (although it can be fun to think about what that solution means), while the other solution is the answer you're looking for.
A: From what you already have,
$$h=5T^2$$
$$x=5(T-1)^2$$
take their ratio
$$\frac{x}{h}=\frac{(T-1)^2}{T^2}=1-\frac{5}{9}=\frac{4}{9}$$
After square root,
$$\frac{T-1}{T}=\frac{2}{3}$$
Thus,
$$T=3$$
A: Note that your assumption that $h - x = \frac59$ is wrong. Observe that the distance travelled by the stone in the last second, which you've denoted by $x$, is actually: $$x = \frac{4h}{9}$$ As a result, what is $h-x?$
A: It took some unknown time $T$ to fall the entire distance $h$ from the top of the tower.
In the last $1$ second the stone fell $\frac59$ of the height of the tower,
that is, the distance it fell was $\frac59h.$
So in the first $T - 1$ seconds it had to fall $\frac49h.$
I look at this and say to myself, the distance fallen is proportional to the square of the time, so
$$ (T-1)^2 = \frac 49 T^2. \tag1 $$
Then I take the square root of both sides. Normally when doing this we have to introduce a $\pm$ sign, but the statement of the problem implies that $T - 1 > 0$ (and therefore $T > 0$), so we only have the positive square root on each side.
I get a linear equation and easily find $T.$
But you might want to be more careful about deriving Equation $(1)$.
I prefer to keep $g$ in the formulas until I need to use its numeric value.
In this case we never need its numeric value, so we never have to make that substitution. The equation for the first $T-1$ seconds is
$$ \frac49 h = \frac g2 (T-1)^2. \tag2$$
For the entire $T$ seconds we have
$$ h = \frac g2 T^2. \tag3$$
Two equations in two unknowns, but very easy to get rid of the unknown $h$ because we can just use Equation $(3)$ as the substitution for $h$ in Equation $(2)$:
$$ \frac49 \left(\frac g2 T^2\right) = \frac g2 (T-1)^2. $$
Cancel the factor of $\frac g2$ on both sides and you get Equation $(1)$.
