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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.

Xander Henderson
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Aditya
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4 Answers4

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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.

Arthur
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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,

$$T=3$$

Quanto
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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 $h-x?$

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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 $(1)$.

David K
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