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

3.2 m

Step-by-step 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 = (v_f^2 -v_i^2)/(2a)=((7.9 m/s)^2-0^2)/(2(9.8 m/s^2))=3.2 m`