Question

5

A marble is dropped from 2.5 m and hits the ground in 0.71 s. What is the final velocity before
it hits the ground?

Answer 1

7.0 m/s

Explanation:

We can solve this using the 2 kinematic equations and you can use either:

`Vf=Vi+at\\\\vf^(2)=vi^(2)+2ad`

So let's use the first one.

Vi = initial velocity

Vf = final velocity

a = acceleration due to gravity (constant 9.8m/s²)

t = time

The marble is dropped so this means that it is in free fall. The initial velocity is always 0m/s.

We are given then the following:

d = 2.5m

t = 0.71s

Vi = 0 m/s

a = 9.8m/s²

So we plug in the values we know and solve:

`Vf = Vi + at\\\\Vf = 0 m/s + (9.8m/s^(2))(0.71s)\\Vf = 6.958m/s \cong 7.0m/s`