Unit 2 Explanation

In this unit I learned about...

Newton's Second Law
Newton's first law states: force is proportional to acceleration and inversely proportional to mass (a=f/m). This means that the more force applied to an object, the more it will accelerate. It also means that as mass increases, acceleration decreases. The equation w = mg (weight equals mass times gravity) tells us how much force is needed to move a particular object. Gravity is equal to 9.8N, but we use 10N to simplify our equations. This means that if we know the mass of something we can also calculate its weight. If an object has a mass of 8kg, then w = (8)(10), w = 80N.

Falling Through the Air with Air Resistance (Skydiving)
fair: force of air resistance
fweight: force of weight (w = mg)
fnet: fweight - fair
When someone first jumps out of an airplane, their velocity is at its lowest point (0 m/s), while their acceleration is at its highest point. When falling, an object's speed increases, and its air resistance increases as well, so speed and air resistance are proportional. This causes its fnet to decrease. Since force is proportional to acceleration, acceleration will decrease as well. Eventually your fair will become equal to your fweight, which means that your fnet will be 0N, so your acceleration will be 0 m/s/s. All of this means that your are at equilibrium (constant/terminal velocity). At this time, velocity is at its highest point, while fnet and acceleration are at their lowest points.

Free Fall - Falling Straight Down
Free fall is when an object falls due to the force of gravity alone. There is no air resistance in free fall, and gravity is the only force acting on the object. This means that the acceleration is always 9.8 m/s/s (10 m/s/s) in free fall, because acceleration is proportional to force, and gravity is the only force present in free fall.

Example: A ball is dropped from a cliff and takes 3 seconds to hit the ground. How far did the ball fall and how fast was it moving when it hit the ground?
To solve for distance...
We use the equation: d = 1/2 at/t
acceleration = gravity, so: d = 1/2 gt/t
plug in what you know: d = 1/2 (10)(3)(3)
solve for distance: d = 1/2 (10)(9)
                                 = 1/2 (90)
                                 = 45 m (don't forget your units!)
To solve for velocity...
We use the equation: v = at
acceleration = gravity, so: v = gt
plug in what you know: v = (10)(3)
                                       v = 30 m/s (units are essential!)

Using these equations, we can calculate the distance and velocity of an object in free fall at any time interval

Throwing Things Straight Up (Free Fall)
So what's the difference between objects being thrown up and an object being dropped? When thrown up, an object has an initial velocity. Our equations d = gt/t and v = gt won't work for objects being thrown up, because these equations assume that the object is starting from rest.

The following images depict (A) free fall, (B) throwing things straight up, and (C) throwing things at an angle (which I will discuss later).

                                            (A)                (B)                                (C)

Remember, as mentioned earlier, the only force present in free fall is gravity (10N) so an object's acceleration is 10 m/s/s. This same rule applies to an object being thrown straight upwards, except the object is decelerating by 10 m/s/s rather than accelerating. The object will continue to decelerate until its velocity is 0 m/s/. When the object's velocity equals 0, then it has reached the top of its path. At this point, it will begin to fall down and accelerate by 10 m/s/s until it again reaches its initial velocity. Picture (B) illustrates what I have just said. Let's practice what we've just learned:

Let's say the ball in picture (B) has an initial velocity of 40 m/s at 0s (initial velocity will always start at 0 seconds). What will its velocity be at 2s?
You may be thinking you need an equation to solve this problem, but you don't! Since the object is being thrown up, it is decelerating by 10m/s every second. In order to find the velocity of this object, we only need to subtract 10m/s for every second that the object decelerates. That means the velocity of the object at 2s is 20m/s.

When the ball is coming back down, the velocities will mirror the velocities on the way up. This means that the velocity at 6s will also be 20m/s, because the object has a constant acceleration of 10m/s/s.

All of this is reiterated in my group's podcast:

<iframe width="560" height="315" src="//www.youtube.com/embed/kJckGQiERvc" frameborder="0" allowfullscreen></iframe>

Falling at an Angle (Free Fall)
Falling at an angle is a type of projectile motion. This means that, along with an increasing vertical velocity, an object also has a constant horizontal velocity. With objects falling at an angle, we must remember that vertical distance determines the time an object is in the air. Falling at an angle is illustrated by a box being dropped out of an airplane and reaching a target below.

In free fall, once this box is dropped it has a constant acceleration of 10m/s/s (g). Let's say it is 125 m off of the ground. How long is the package going to be in the air? Remember time is determined by vertical distance. We can solve for time by using the distance equation:
d = 1/2 gt^2
125 = 1/2 (10)(t^2)
125 = (5) (t^2)
25 = t^2
5s = t

How much farther back from the target must the plane drop the box? This is just a different way of saying, what is the horizontal distance from the target? Let's say the package is moving at 90 m/s in the horizontal direction. We can use the velocity equation to solve for distance:
v = d/t
90 = d/5

450m = d

Throwing Things Up at an Angle (Free Fall)
This is another type of projectile motion. Just as when an object is thrown straight upward, objects thrown up at an angle have an initial velocity. In addition, they also have a constant horizontal velocity (like things falling at an angle). This type of motion is depicted in image (C).

                                            (A)                  (B)                          (C)
To find the resultant velocities of the horizontal and vertical velocities at a given time interval, we use the equation a^2 + b^2 = c^2
We use the same equations used for things falling at an angle to find distance and time.


My problem solving skills, effort, and learning...

In this unit I was diligent throughout my work. I participated in class, and completed my homework to the best of my ability, although sometimes my workload prohibited me from doing it fully. In group activities I happily gave and received help on certain subjects. I have completed all of my blog posts on time and have featured the required criteria. I feel that I have been very persistent throughout my work, never hesitating to ask questions in class and during conference period. I am learning to take the information we've learned in class and apply it to more difficult problems, as well as work effectively within a group to accomplish challenging assignments. My goal for the next unit is to complete all of my homework to its fullest and continue engaging in class.

Connections...

It is useful for me to know about falling at an angle, because pilots must know about this when dropping relief packages. I could also use what I learned about falling straight down to calculate how far a jump might be and whether or not it would be safe.