Problem — From final speed to height

Problem

Inverse: given the final speed, find $h$. From what height would a stone have to fall vertically, starting from rest, to reach the ground with the speed of a car travelling at $90\;\text{km/h}$?

Solution

Speed conversion. Recalling that $1\;\text{km/h} = 1/3{,}6\;\text{m/s}$:

$v = \frac{90}{3{,}6} = 25\;\text{m/s}$

Speed–height relation. In free fall from rest the arrival speed is linked to the height by the relation without time (with $a = g$):

$v^2 = 2\,g\,h$

Finding the height. We isolate $h$ and use $g = 9{,}81\;\text{m/s}^2$:

$h = \frac{v^2}{2\,g} = \frac{25^2}{2\cdot 9{,}81} = \frac{625}{19{,}62}\;\text{m}$

$\ev{h \approx 31{,}9\;\mathrm{m}}$

About a $10$-storey building: it is surprising how high that is, but it explains why a fall of just a few metres already reaches dangerous speeds.

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Topics: Cinematica Concepts: Caduta libera