Free fall in two dimensions

Problem

Free fall in 2D. A stone is thrown horizontally at $15\;\text{m/s}$ from a building $20\;\text{m}$ high. Calculate: (a) the flight time; (b) the horizontal distance at the ground. The vertical component does not depend on the horizontal one.

Solution

The motion splits into two independent motions along the Cartesian axes:

• along $x$ (horizontal): uniform motion with constant speed $v_0 = 15\;\text{m/s}$;
• along $y$ (vertical): free fall from rest, with acceleration $g$.

(a) Flight time. This depends only on the vertical motion: the stone must descend by $h = 20\;\text{m}$ starting with zero vertical speed. From $h = \tfrac{1}{2}g\,t^2$:

$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \cdot 20}{9{,}81}} = \sqrt{4{,}077}\;\mathrm{s} \approx 2{,}02\;\mathrm{s}$

The horizontal speed does not affect this time at all: the stone hits the ground at the same instant it would if it were simply dropped.

(b) Horizontal distance (range). Along $x$ the motion is uniform, so the distance is speed times time:

$x = v_0 \cdot t = 15 \cdot 2{,}02 \approx 30{,}3\;\mathrm{m}$

$\ev{t \approx 2{,}02\;\mathrm{s}, \quad x \approx 30{,}3\;\mathrm{m}}$

Links

Topics: Cinematica · Vettori Concepts: Moto parabolico · Caduta libera Methods: Scomposizione in componenti cartesiane