Ranking fall times

Problem

Ranking: fall times. Four balls fall in a vacuum from heights: (A) $h = 0{,}5\;\text{m}$; (B) $h = 2\;\text{m}$; (C) $h = 5\;\text{m}$; (D) $h = 20\;\text{m}$. Rank the fall times from shortest to longest.

Solution

For free fall from rest, the time depends only on the height:

$h = \frac{1}{2}\,g\,t^2 \quad\Longrightarrow\quad t = \sqrt{\frac{2h}{g}}$

Since $t$ increases monotonically with $h$, a greater height corresponds to a longer fall time: the order of the times follows the order of the heights. Let us compute the values with $g \approx 9{,}81\;\mathrm{m/s^2}$:

$t_A = \sqrt{\tfrac{2 \cdot 0{,}5}{9{,}81}} \approx 0{,}32\;\mathrm{s}, \quad t_B = \sqrt{\tfrac{2 \cdot 2}{9{,}81}} \approx 0{,}64\;\mathrm{s}$

$t_C = \sqrt{\tfrac{2 \cdot 5}{9{,}81}} \approx 1{,}01\;\mathrm{s}, \quad t_D = \sqrt{\tfrac{2 \cdot 20}{9{,}81}} \approx 2{,}02\;\mathrm{s}$

Note that the time grows only as $\sqrt{h}$: quadrupling the height doubles the time. For example, going from $B$ ($2\;\text{m}$) to $D$ ($20\;\text{m}$) the height increases $10$-fold, but the time only grows by a factor $\sqrt{10} \approx 3{,}2$.

$\ev{A < B < C < D \quad (0{,}32 < 0{,}64 < 1{,}01 < 2{,}02\;\mathrm{s})}$

Collegamenti

Argomenti: Kinematics Concetti: Free fall Competenze: Ranking reasoning