Question

A pencil rolls horizontally of a 1 meter high desk and lands .25 meters from the base of the desk. How fast was the pencil rolling when it left the desk?

Answer 1

Answer: 0.55 m/s

Step-by-step explanation:

This situation is related to projectile motion (also called parabolic motion), where the main equations are as follows:

`x=V_(o) cos\theta t` (1)

`y=y_(o)+Vo sin \theta t + (g)/(2)t^(2)` (2)

Where:

`x=0.25 m` is the horizontal displacement of the pencil

`V_(o)` is the pencil's initial velocity

`\theta=0\°` since we are told the pencil rolls horizontally before falling

`t` is the time since the pencil falls until it hits the ground

`y_(o)=1 m` is the initial height of the pencil

`y=0` is the final height of the pencil (when it finally hits the ground)

`g=-9.8m/s^(2)` is the acceleration due gravity, always acting vertically downwards

Begining with (1):

`x=V_(o) cos(0\°) t` (3)

`x=V_(o)t` (4)

Finding
`t` from (2):

`0=1 m+ (-9.8m/s^(2))/(2)t^(2)` (5)

`t=\sqrt{(-2y_(o))/(g)}` (6)

Substituting (6) in (4):

`x=V_(o)\sqrt{(-2y_(o))/(g)}` (7)

Isolating
`V_(o)`:

`V_(o)=\frac{x}{\sqrt{(-2y_(o))/(g)}}` (8)

`V_(o)=\frac{0.25 m}{\sqrt{(-2(1 m))/(-9.8m/s^(2))}}` (9)

Finally:

`V_(o)=0.55 m/s`