1) The object is released from rest at a height of -40.8 m above a reference point with an acceleration due to gravity of -9.81 m/s^2.
2) Using the kinematic equations, the final velocity vy is calculated to be -28.3 m/s.
3) As the object falls, it initially accelerates downward at g but then reaches a smaller acceleration a < g as drag force opposes the weight, until terminal velocity Vt is reached where drag equals weight and acceleration is zero.
Putter King Education - Physics (Level 2)putterking
Here you can learn all about the physics concepts that are hidden in miniature golf. Visit www.putterking.com for more info.
Level 2 - Princess
Area of focus: force and motion
Topics covered:
> Force
> Gravity
> Law of Universal Gravitation
> Mass vs. weight
> Newton’s First Law of Motion
> Newton’s Second Law of Motion
> Newton’s Third Law of Motion
Our project in physics (IV-Einstein) about waves: It's nature, types, parts and measures.
I apologize for the ugly font, I used different font styles that are not available on all computers since they are downloaded from the internet.
Putter King Education - Physics (Level 2)putterking
Here you can learn all about the physics concepts that are hidden in miniature golf. Visit www.putterking.com for more info.
Level 2 - Princess
Area of focus: force and motion
Topics covered:
> Force
> Gravity
> Law of Universal Gravitation
> Mass vs. weight
> Newton’s First Law of Motion
> Newton’s Second Law of Motion
> Newton’s Third Law of Motion
Our project in physics (IV-Einstein) about waves: It's nature, types, parts and measures.
I apologize for the ugly font, I used different font styles that are not available on all computers since they are downloaded from the internet.
Physics Investigatory Project on Fluid Mechanicsashrant
A project on the relation between the coefficient of viscosity and the SAE grading system of engine oils.
Complete with a postulated equation to calculate the viscosity of a liquid at ant given temperature.
1. Sy = -40.8 m
ay = -9.81 ms-2
Reference
uy=0
Point
Find vy.
40.8 m
v = u + at
s = ut + ½ at2
v2 = u2 + 2 as
Taking upwards
as positive vy=-28.3m/s
flipperworks.com
2. Qualitative Description of Motion
Positive At A, object is first released
Velocity
Uy = 0 ms-1 no air resistance
Resultant force, Fnet is equal to the
weight W, which is the only force
acting on the object.
Acceleration = g = 9.81m/s
Weight (W =mg)
Fnet = mg
Object is accelerating. flipperworks.com
3. V / ms-1
Vt
C Qualitative Description of Motion
B
At B, object is moving downwards
A t/s
•As motion is downward, the drag
force must be upwards.
Drag Force(Fd) •Object accelerates with smaller
acceleration. acceleration, a < g
•Downward velocity still increasing
Velocity
Weight (W=mg) Fnet = mg - Fd flipperworks.com
4. V / ms-1
C
Qualitative Description of Motion
Vt
At C, object reaches terminal velocity
B
A t/s •As velocity increases, the drag force
increases as well.
•Drag force equal and opposite to the
Drag Force
weight: no resultant force
•Acceleration is zero
•Velocity stops increasing terminal
velocity (Vt)
Velocity
(Vt) Fnet = mg – Fd = 0
Weight flipperworks.com
acceleration, a = 0