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

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)

$y=y_{o}+Vo sin \theta t + \frac{g}{2}t^{2}$ (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):

$0=1 m+ \frac{-9.8m/s^{2}}{2}t^{2}$ (5)

$t=\sqrt{\frac{-2y_{o}}{g}}$ (6)

Substituting (6) in (4):

$x=V_{o}\sqrt{\frac{-2y_{o}}{g}}$ (7)

Isolating $V_{o}$:

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

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

Finally:

$V_{o}=0.55 m/s$