This
unit we went into a lot more complex concepts than the first unit. This unit we
learned about four specific concepts; free fall, throwing objects straight into
the air, projectile motion, throwing objects at an angle, and falling through
the air.
To
start off the unit we began with Newton’s second law of motion. This states
that force causes acceleration therefore force is directly proportional to
acceleration and acceleration is inversely proportional to mass. The formula
for this law is (a=fnet/m).
In
the beginning we learned various formulas, once of which I just listed above.
Well let’s say you’re given the mass of the object but you’re not given the
weight of the object. Here, you would use (weight=(mass)x(gravity)). An example
of this sort of problem would be “An object is given a mass of 50 kg. What is
the objects weight?” You could
solve this simply by using this formula (w=mg). You would plug in 50 for m and
10 for g because gravity will always equal 9.8m/s^2 or the rounded version that
we use of 10m/s^2. After plugging these numbers in you would get 500N. Remember
all weight must be in Newtons for physics.
Next,
we learned about the concept of free fall. Free fall is when an object falls
due to the effect of gravity alone. There is no other force acting the object
besides gravity. This also dismisses air resistance from the equation. Here we
are introduced to a few equations (d=at^2 or d=gt^2 and v=at or v=gt). In a
free fall equation we can say that acceleration is equal to gravity because
during free fall, it goes at a constant rate of 10m/s^2 and because the only
force acting the object is gravity, that means the only force acceleration the
object is gravity. So, gravity would equal (10m/s^2) as well. Here is a good
example of free fall; “You drop an object from the top of the building and it
takes 3 seconds to hit the ground. How high is the building and how fast was
the object moving when it hit the ground. These were easy to solve with the new
formulas we had been given. For the how far equation you would use
(d=1/2gt^2). You would then
evaluate this equation and end up with (d=1/2(10)(3)^2). The answer would be 45
m tall building. For the second part of the question for the how fast answer
you would just use the (v=gt) equation. (v=10(3)). Therefore it would be 30m
per second. Going into more detail about objects in free fall, because the only
force acting on an object in free fall all objects will hit the ground at the
same time no matter what their weight is, as long as you are neglecting air
resistance. Therefore when a penny and a feather are placed inside a vacuum
tube, both will hit the ground at the same time because they are released from
the same distance at the same time and there is no air resistance in this
equation therefore the only force acting upon these objects is the force of
gravity. To test free fall and better understand it, we did a lab where we
estimated to calculate the actual height of 3rd Anderson.
After
free fall we moved on to throwing things straight up into the air. Most of the
properties of these go back to similar ones of that of free fall. We learned
that the initial velocity of an object thrown into the air decreases by 10m/s^2
until it reaches the top of its path where it is still accelerating at 10m/s^2
but its velocity is 0m/s. This is known as its “hang time”. Some objects have
more hang time than others based on their vertical distance. The vertical
distance controls the amount of time an object will spend in the air. As the ball continues back down towards
the earth, again, it has an acceleration of 10m/s^2 and an increasing velocity
before it hits the ground. An example of this would if an object were thrown
straight into the air with an initial velocity of 40m/s (neglecting air
resistance) how long will the ball spend in the air? What is the vertical
distance of the ball? You would use the formulas (d=1/2gt^2 and v=gt) to find
that the ball was in the air for 4seconds and it traveled about 80m to reach
the top of its path.
After
mastering the concept of throwing objects directly into the air, we moved on to
projectile motion. This is where the new concept of a horizontal and a vertical
path. When an object, let’s say a package, is dropped from a plane moving
through the sky, the path of this object may appear to be moving straight down
if you were looking at from the plane, however, the real path of this package
is more of a parabola, or diagonal line. Therefore, we have to take into
consideration the vertical path and its horizontal path. The vertical velocity
of the ball will continuously increase as it moves downward and it will have a
constant acceleration. The vertical distance controls the time in which the
ball will spend in the air. Again, we use the formulas we’ve used in the past
to figure out all of these factors (d=1/2gt^2 and v=gt). Now however to figure
out the horizontal velocity, which will remain constant and how far the ball
moved in the horizontal direction, we use the formula (v=d/t). Time is constant
for both the horizontal and vertical directions. Even if someone jumps father
out, than the distance a person jumps upward, they person jumping farther out
will land first because they don’t have as great of a vertical height. Therefore,
when a bullet and its shell hit the ground at the exact same time even though
the bullet travels father down field its because they both travel the same
vertical distance, so when the two leave the gun the only force acting on these
objects is the force of gravity. So the bullet may have a higher acceleration
and the shell may have a higher velocity but the force of gravity pulls
downward at the same exact time. This also answers the question “If a monkey
lets go after a dart is fired at him, ill he avoid being hit? The answer is no,
he will still be hit because the dart and the monkey started at the same height
therefore; gravity will both pull them towards the ground at the exact same
rate.
The
next part of this unit we learned was throwing things up at an angle. The
vertical velocity controls the time an object is in the air. Because the
gravity causes the velocity to decrease at 10m/s every second in the middle of
an objects path being thrown into the air will always have a velocity of 0m/s.
Now that we are throwing objects at an angle, we need to figure out the actual
diagonal path it’s taking. We can do this by using the formula (a^2 + b^2 =
c^2) or memorizing the most important triangles; 3,4,5, 30, 40, 50 or 1,1,
1.41,etc.
The
last section we worked on was falling through the air. Air resistance becomes
stronger as you go faster. The force of air resistance increases causing the
speed of an object to increase. Acceleration is decreased because net force is
decreased. Velocity increases. To find the acceleration when an object is
falling through the air you use the formula a=fnet/m. The only way for fair to
decrease is for the skydiver to slow down because when the skydiver opens the
parachute it causes everything to be off balance. Air resistance is much
greater than its velocity. Therefore the object has to increase its velocity to
decrease the air resistance an math its air resistance reaching terminal
velocity.
