Question

What vertical distance Δy does a free-falling particle travel from the moment it starts to the moment it reaches a speed of 7.9 m/s if it starts from rest? Work out your solution using one of the equations for vertical motion with constant acceleration, specifically, v2f=v2i+2aΔy, where vi and vf are the particle’s initial and final speeds, respectively, and a is the particle’s acceleration.

Answer 1

Answer:

3.2 m

Explanation:

The equation to use to solve this problem is:

$v_f^2 = v_i^2 + 2 a \Delta y$

where

$v_f$ is the final velocity

$v_i$ is the initial velocity

a is the acceleration

$\Delta y$ is the distance covered

For the particle in free-fall in this problem, we have

$v_i = 0$ (it starts from rest)

$v_f = 7.9 m/s$

$g=9.8 m/s^2$ (acceleration due to gravity)

By re-arranging the equation, we can find the distance travelled:

$\Delta y = \frac{v_f^2 -v_i^2}{2a}=\frac{(7.9 m/s)^2-0^2}{2(9.8 m/s^2)}=3.2 m$