Unit Summary Blog Post 

Part A)

In this unit I learned about …

Newton’s Second Law

Newton’s Second Law is easily stated as a= F/m. When you break this down, we can see that force is directly proportional to acceleration (F~a) which means that that when the force increases or decreases, acceleration does the same. The second part states that acceleration is inversely proportional to mass (a~1/m) which means that as the acceleration increases or decreases, the mass does the opposite. *Also, if you ever need to find the weight of an object, use the formula w=mg.

Example:

If one person is providing 100N of force to an object and the acceleration is 2m/s2, what is the new acceleration of your force increased to 200N?

a=F/m

2m/s2=100/m

m=50

a=200/50

a=4m/s2

Free Fall Straight Down (with no air resistance)

Free fall is when an object is falling through the air with no air resistance. This means that the only force working on the object is gravity. Gravity is technically equal to 9.8m/s2, but for simpler purposes we mainly used 10m/s2 in class. The acceleration of an object in free fall is always a constant 10m/s2. There are two formulas to calculate the vertical free fall of an object. The formula for calculating distance is d=1/2gt2 and the formula for calculating velocity is v=gt. Below is a podcast that Mark and I created to help us to understand free fall better.

Example: 

If I fell off of my balcony, and it took me 5 seconds to land in the pool below, how high was my balcony and how fast was I falling just before I hit the pool?

d=1/2gt2

d=1/2g(5)2

d=1/2g(25)

d=5(25)

d=125m

And if you were trying to calculate the velocity at which I was falling:

v=gt

v=10(5)

v=50m/s

Free Fall At an Angle (with no air resistance)

Free fall at an angle is very similar to regular free fall, except now we have a constant horizontal velocity. Free falling at an angle could be anything from a plane dropping a package below as it continues to fly, to a person running straight off of a cliff. *An important thing to keep in mind is that even when that package is released from the plane or the man finally reaches the edge of the cliff, the horizontal velocity does not change. Meaning that when the package hits the ground, it would still be directly below the plane regardless of where it was released.

Example:

You throw a ball off of a cliff that is 80m high with an initial horizontal velocity of 5m/s. Calculate the time that the ball spent in the air and how far away from the cliff it landed.

d=1/2gt2

80=1/210(t2)

80=5(t2)

80/5=t2

16=t2

t=4 seconds

What is the diagonal velocity (hypotenuse of the special right triangle)?

a2+b2=c2

(5)2+(10)2=c2

25+100=c2

square root of 125 = 11.18m

Throwing Things Straight Up (Free Fall)

When throwing an object straight up into the air (neglecting air resistance) we can be sure that the acceleration will always be 10m/s2. The object will have an initial velocity of whatever force you throw it up with and continues to decrease by 10m/s until the velocity reaches 0m/s and the ball is at the top of its path. Once the ball reaches the top of its path, it will begin to free fall back down. This portion of the free fall is no different from just dropping an object from the same height. 

Example: 

A ball is thrown straight up with an initial speed of 40m/s. How long does it take for the ball to reach the top of its path? = 4 seconds

How long is the ball in the air? = 8 seconds

How high was the ball at the top of its path? 

d=1/2gt2 

d=1/2g(4)2

d=5(16)

d=80m

How high was the ball at 2 seconds?

d=1/2gt2

d=1/2g(2)2

d=5(4)

d=20m

80-20

distance at 2 seconds= 60m

Throwing Things Up At An Angle 

When throwing things up at an angle (projectile motion) we must remember the two different sets of formulas.

Horizontal-

d=vt

v=d/t

Vertical-

d=1/2gt2

v=gt

Example:

A man throws a ball with an initial velocity of 100m/s, and the ball is in the air for 10 seconds. Using the 1, 1, 1√2 triangle, (there is also a 3,4,5 triangle) we can see that the actual velocity of the ball is 100√2, which = 141m/s. 

The veritcal distance of the ball-

d=1/2gt2

d=1/2g(10)2

d=5(100)

d=500m

The horizontal distance of the ball-

d=vt

d=100(10)

d=1000m

Falling Through Air WITH Air Resistance (Skydiving)

Finally, air resistance comes into play when we are talking about skydiving. When you jump out of a plane and begin to fall down to the earth, there are two forces to be aware of. Your downward force is called F-weight. As you fall, you are gaining speed, meaning that your your F-weight is increasing, even though your acceleration is decreasing. Since your speed is increasing downwards, this causes your upward force to increase as well. This upward force is called F-air. In the first stages of the fall, your F-weight is much greater than your F-air. Velocity is directly proportional to air resistance, resulting in the increase of F-air until you reach your first terminal velocity. At terminal velocity, your velocity is remaining constant while your acceleration is at 0m/s2.

Opening the parachute:

When you first open the parachute, your magnitude automatically increases in the upward direction. Although, as you continue to fall, your magnitude decreases in the upwards direction until your F-weight and F-air equal out again and you reach your second terminal velocity. This terminal velocity is similar to the first but it is much, much slower.

*Remember* that net force and acceleration are directly proportional to each other. Also, keep in mind that the forces of air resistance are equal and opposite to weight.

Part B)

This all connects to our everyday lives without us noticing. Tossing a ball to your friend during afternoon activities or even just accidentally dropping your phone are all actions that can be better understood with physics. Each of the examples I gave above were real life ways of looking at free fall and newton's second law. I hope that they help you to gain a less complex understanding of unit 2!