Stone from the cliff

Problem

A stone is dropped from a cliff and takes $3{.}0\;\text{s}$ to reach the ground. Neglecting air resistance, calculate the height of the cliff and the impact velocity.

Solution

Setup. The stone is “dropped”, so the initial velocity is zero ($v_0 = 0$). Neglecting air, this is free fall: uniformly accelerated motion with acceleration $g \approx 9{.}81\;\text{m/s}^2$ directed downward. We take the axis oriented downward with origin at the starting point.

Height of the cliff. The distance covered in free fall from rest is

$h = \frac{1}{2} g\, t^2 = \frac{1}{2}\cdot 9{.}81\cdot (3{.}0)^2 = \frac{1}{2}\cdot 9{.}81\cdot 9{.}0 \approx 44{.}1\;\text{m}$

$\ev{h \approx 44\;\mathrm{m}}$

Impact velocity. With $v_0 = 0$ the velocity grows linearly with time:

$v = g\, t = 9{.}81\cdot 3{.}0 \approx 29{.}4\;\text{m/s}$

$\ev{v \approx 29{.}4\;\mathrm{m/s} \approx 106\;\mathrm{km/h}}$

Check. The same velocity is obtained from the time-free relation $v = \sqrt{2 g h} = \sqrt{2\cdot 9{.}81\cdot 44{.}1} \approx 29{.}4\;\text{m/s}$, consistent with the previous result. As always in free fall, height and impact velocity do not depend on the mass of the stone.

Collegamenti

Argomenti: Kinematics Concetti: Free fall · Uniformly accelerated motion