Answer: 0.55 m/s
Explanation:
This situation is related to projectile motion (also called parabolic motion), where the main equations are as follows:
[tex]x=V_{o} cos\theta t[/tex] (1)
[tex]y=y_{o}+Vo sin \theta t + \frac{g}{2}t^{2}[/tex] (2)
Where:
[tex]x=0.25 m[/tex] is the horizontal displacement of the pencil
[tex]V_{o}[/tex] is the pencil's initial velocity
[tex]\theta=0\°[/tex] since we are told the pencil rolls horizontally before falling
[tex]t[/tex] is the time since the pencil falls until it hits the ground
[tex]y_{o}=1 m[/tex] is the initial height of the pencil
[tex]y=0[/tex] is the final height of the pencil (when it finally hits the ground)
[tex]g=-9.8m/s^{2}[/tex] is the acceleration due gravity, always acting vertically downwards
Begining with (1):
[tex]x=V_{o} cos(0\°) t[/tex] (3)
[tex]x=V_{o}t[/tex] (4)
Finding [tex]t[/tex] from (2):
[tex]0=1 m+ \frac{-9.8m/s^{2}}{2}t^{2}[/tex] (5)
[tex]t=\sqrt{\frac{-2y_{o}}{g}}[/tex] (6)
Substituting (6) in (4):
[tex]x=V_{o}\sqrt{\frac{-2y_{o}}{g}}[/tex] (7)
Isolating [tex]V_{o}[/tex]:
[tex]V_{o}=\frac{x}{\sqrt{\frac{-2y_{o}}{g}}}[/tex] (8)
[tex]V_{o}=\frac{0.25 m}{\sqrt{\frac{-2(1 m)}{-9.8m/s^{2}}}}[/tex] (9)
Finally:
[tex]V_{o}=0.55 m/s[/tex]
