Free Fall
Free Fall - when objects fall due to the acceleration of gravity only
-no air resistance
-weight doesn't get factored in-when falling, the object increases by 10m/s every second
To find how fast the object is moving, use the formula --> v = gt
To find how far the object has gone, use the formula --> d = 1/2gt^2
Example problem:
A ball is dropped from a cliff, and it took the ball 3 seconds to hit the ground.
How high up is the cliff? How fast is the ball moving?
d=1/2gt^2 v=gt
=1/2(10)(3^2) =(10)(3)
=(5)(9) =30m/s
=45m
Projectile Motion
constant a constant v
(vertical distance height) (horizontal distance)
d=1/2gt^2 v=d/t
v=gt d=vt
*vertical distance can
only be calculated if
the object starts at rest
(0m/s) so only if the object
is in free fall
A plane at a height of 125m is going 90m/s drops a package. How long is the package going to be in the air?
*the only thing controlling an object's time in the air is the vertical distance (hang time)
d=1/2gt^2
125=1/2(10)(t^2)
125=(5)(t^2)
t^2=25
t=5
How far back will the plane need to drop the package in order for it to hit the target?
*We're looking for the horizontal distance
d=vt
=(90)(5)
=450m
Velocity in the horizontal distance is treated differently than how we previously treated the vertical distance. The horizontal velocity stays constant, unlike the vertical velocity that is increasing by 10m/s
every second.
Throwing Things At An Angle
Someone hits a baseball, and this is the shape the ball took in the air. The ball is in the air for 4 seconds and goes a total distance of 120m. At the top of the ball's path, it is only moving with horizontal velocity. What is its horizontal velocity?
v=d/t
If we didn't already know, and wanted to find =120/4
how far downhill the ball would land... =30m/s
d=vt
=(30)(4)
=120m
a^2+b^2 = c^2
(20^2) + (30^2) = c^2
400+900 = c^2
c^2=1300
c=36m/s
Newton's 2nd Law
Force causes acceleration.
-force is proportional to acceleration (if force decreases, then acceleration decreases, and if force increases, then acceleration increases)
mass increase --> acceleration decreases
mass decrease --> acceleration increases
acceleration is directly proportional to force --> a~F
acceleration is inversely proportional to mass --> a~1/m
Newton's 2nd Law --> a = F/m
(this is true because a~F and a~1/m)
The law in words --> Acceleration is directly proportional to force, and acceleration is inversely proportional to mass.
You are pushing a 10kg box to the right with a force of 50N and your friend is pushing the box in the opposite direction with 10N. What is the acceleration of the box? What direction is the box accelerating?
a = F/m
= (50-10) / 10
= 40 / 10
= 4m/s^2 in the right direction
The law in words --> Acceleration is directly proportional to force, and acceleration is inversely proportional to mass.
a = F/m
= (50-10) / 10
= 40 / 10
= 4m/s^2 in the right direction
What is the weight of a 10kg box with the upward force of 50N and the force of gravity in the opposite direction?
*weight (net force) = (mass) (gravity)
w = mg
=(10)(10)
=100N
Newton's 2nd Law Lab
The hanging weight will apply the force that causes acceleration.
If the mass of the cart increases, but the force (hanging weight) remains the same, then the acceleration will decrease because mass and acceleration are inversely proportionate.
If we keep the hanging weight the same, then the net force on the system will remain constant.
If we take mass from the cart and move it to the hanger, the acceleration will increase because the acceleration is directly proportional to force, and the weight of the hanger is the force that accelerates the system. *Doing this is still keeping the mass of the system constant*
BUT we can't just simply add mass to the hanger from an outside force, because that would be changing the mass of the system AND the force.
If the net force on a body is zero, the acceleration of that body is also zero because force and acceleration are directly proportionate.
--> If the net force on a body is constant, the acceleration is also constant.
a = F/m looks like y = mx+b
so... whatever is kept constant in the experiment is the slope in the equation of line.
If the mass was kept constant --> a = (1/m)(F)
If the force was kept constant --> a = (F)(1/m)
To determine if your date confirms Newton's 2nd Law, your generated slope needs to be within 10% of the given slope.
For example, if the given line is y = 0.5531x and your generated slope is .49 then your date confirms Newton's 2nd Law because .49 is within 10% of .5531
A person with a weight of 100N is skydiving. The Fair is 20N.
*remember that if speed increases, air resistance also increases
Fnet = Fweight - Fair
Fnet = 100N - 20N
Fnet = 80N
*If Fnet decreases, then the acceleration also has to decrease because force and acceleration are directly proportional. (the weight is the force)
As the person is falling, their speed is increasing, thus increasing their air resistance.
t = 0
Fweight = 100N
Fair = 20N
t = 1
Fweight = 100N
Fair = 30N
t = 2
Fweight = 100N
Fair = 40N
*Acceleration is decreasing, but speed still is still increasing.
This person has reached a point in skydiving called terminal velocity, which is how fast an object can possibly go.
Fnet = Fweight - Fair
Fnet = 100-100
Fnet = 0N
Terminal velocity is achieved if...
-Fnet = 0N
-a = 0m/s^2
-the object is at equilibrium
-the object is moving at constant velocity
Thrown Straight Up
