We learned about three different types of free fall: free fall straight down, free fall thrown upward, free fall with projectile motion, and free fall straight down with air resistance.
1) The simplest form of free fall would be, for example, a ball was dropped off of a building. In this situation, air resistance is negligible and the object is falling from rest. Free fall is when objects fall due to the acceleration of gravity only.
Say you were on a hike. You come across a cliff. You want to know how far down the drop is. In order to do this, you could drop a rock off of the cliff and time how long it took to make the fall. In order to calculate the hight of this cliff, you need to know a formula: d=1/2gt^2. In this formula, g or the force of gravity, will theoretically always be 10 (but in the real world it is 9.8). Say it took the rock 8 seconds to fall. You would say d=1/2x10x8^2. If you solve this equation, you would find out that the distance, or hight of the cliff is 320m.
If you want to know how fast the rock, or whatever object you drop from whatever hight is moving at a certain time, you need to know the equation: v=gt and remember that because the force of gravity is 10, our velocity will increase by 10m/s every second. So say you want to know how fast the object was moving after 5 seconds of falling. You would simply multiply 10 (gravity) by 5 (time) and you would get the velocity at that time. In this scenario, the velocity would be 50 m/s.
Say you were on a hike. You come across a cliff. You want to know how far down the drop is. In order to do this, you could drop a rock off of the cliff and time how long it took to make the fall. In order to calculate the hight of this cliff, you need to know a formula: d=1/2gt^2. In this formula, g or the force of gravity, will theoretically always be 10 (but in the real world it is 9.8). Say it took the rock 8 seconds to fall. You would say d=1/2x10x8^2. If you solve this equation, you would find out that the distance, or hight of the cliff is 320m.
If you want to know how fast the rock, or whatever object you drop from whatever hight is moving at a certain time, you need to know the equation: v=gt and remember that because the force of gravity is 10, our velocity will increase by 10m/s every second. So say you want to know how fast the object was moving after 5 seconds of falling. You would simply multiply 10 (gravity) by 5 (time) and you would get the velocity at that time. In this scenario, the velocity would be 50 m/s.
2) You are watching a soccer game. The goalie catches the ball and decides to punt it down the field. She tosses the ball up, waits for it to come back down, and then kicks it. You can figure out how high the ball goes, how long it is in the air, and how fast it is moving at any time. As you can see in the picture, this speed will decrease by 10m/s every second so by drawing out what we know, we can figure out the aforementioned three things just by knowing the starting velocity or the time in the air.
If we want to know high the ball was at the top of its path before it started to fall, we use our trusty d=1/2gt^2 formula. So in this case, we have figured out that the ball was in the air for 4 seconds before reaching the top of its path. If we plug that number in for t, we find that it reached a hight of 80m. Lets say we want to know the balls hight at 2 seconds. In order to find this, we need to find the total hight as we did before (pictured in green) and subtract this from the distance pictured in blue. This equals the orange hight, or the hight at 2 seconds.
If we want to know high the ball was at the top of its path before it started to fall, we use our trusty d=1/2gt^2 formula. So in this case, we have figured out that the ball was in the air for 4 seconds before reaching the top of its path. If we plug that number in for t, we find that it reached a hight of 80m. Lets say we want to know the balls hight at 2 seconds. In order to find this, we need to find the total hight as we did before (pictured in green) and subtract this from the distance pictured in blue. This equals the orange hight, or the hight at 2 seconds.
3) Once the goalie kicks the ball, it is propelled foreword and shoots down the field. This is an example of projectile motion. The goalie kicks the ball 50m down the field at a 45 degree angle and the ball stays in the air for 4 seconds. You can figure out how hard she kicked the ball. As you can see in the picture, you can predict where the ball would be every second and how fast it would be going at each second. If we look at our picture, we can see that the ball was kicked at a vertical speed of 20m/s at 0seconds. Remember that its horizontal velocity is constant. In order to find it, we just plug our numbers into the formula: v=d/t. In order to figure out how fast the ball was actually moving at any given time, we draw out a picture like this: The actual speed will be the hypotenuse. In order to solve for this, we just use the pythagorean theorem: a^2+b^2=c^2, or one of our special triangles: 3, 4, 5, or x, x, xsqrt2. Remember that the square root of 2 is 1.41. So we can find out that in order for the soccer ball to land 50m away, the goalie must kick the ball at about 22 m/s. We can figure out how long the ball will be in the air, how fast the ball will be at the top of its path, and how far away it will land. Remember that the vertical velocity is the main component. In summary, the big formulas to remember are: vertical: d=1/2gt^2, v=gt horizontal: d=vt, v=d/t.
So lets say that for some strange reason, the goalie who is holding a ball sprouts wings and begins to fly foreword at a speed of 90m/s 125 meters above the ground. She wants to make a goal by dropping the ball into it. The ball has a constant vertical acceleration and a constant horizontal velocity. First lets find out how long the ball will be in the air by using d=1/2gt^2. We should get 5s. Remember we have to drop the ball very early because the ball is moving foreword at 90m/s and will continue to because of inertia. In order to find how far away to drop the ball in order to make a goal, we use the formula v=d/t and find out that horizontal distance is 450m. In order to find where the ball would be in the vertical direction each second, we use d=1/2gt^2. In order to see the actual estimated path, just find where the horizontal and vertical would meet up.
So lets say that for some strange reason, the goalie who is holding a ball sprouts wings and begins to fly foreword at a speed of 90m/s 125 meters above the ground. She wants to make a goal by dropping the ball into it. The ball has a constant vertical acceleration and a constant horizontal velocity. First lets find out how long the ball will be in the air by using d=1/2gt^2. We should get 5s. Remember we have to drop the ball very early because the ball is moving foreword at 90m/s and will continue to because of inertia. In order to find how far away to drop the ball in order to make a goal, we use the formula v=d/t and find out that horizontal distance is 450m. In order to find where the ball would be in the vertical direction each second, we use d=1/2gt^2. In order to see the actual estimated path, just find where the horizontal and vertical would meet up.
4)In this unit we also learned about Newtons Second Law. The law states that acceleration is directly proportional to force and inversely proportional to mass. This statement can also be written as a formula: a=F/m or a=Fx1/m. This means that if acceleration increases, force would increase or if acceleration decreased, force would decrease also. This also means that if acceleration increases, mass decreases or if acceleration decreased, mass would increase. In the real world, Newtons Second Law can be seen in a person pushing a box. If the box was light (or had a small mass) it would be easier to move, or accelerate and if the box was heavy (or had a large mass) it would be more difficult to move or accelerate. If you had a box and pushed it just a little bit (with a small amount of force) it would not accelerate quickly. If you pushed the same box with a lot of force it would accelerate quickly.
5) We did a lab to demonstrate these concepts and illustrate how acceleration depends on force and mass. In this lab, we had a cart on a track with a string attached. The string ran over a pulley to a hanging weight below. In this example, the hanging weight applied the force that caused the acceleration. We needed to find all of the components in our formula: a=F/m. In this case, the acceleration is basically how fast the cart goes. To find the mass, we added up the masses of the cart and hanger to find the total mass of the system. To find the force, we found the weight of the hanger by using the formula w=mg (weight equals mass times force of gravity) and kept this constant throughout the experiment. In .10 kg increments we added more masses onto the cart therefore changing the the total mass of the system. We found that as the mass of the cart increased, the acceleration decreased. In our next part of the experiment, we kept the total mass constant, but moved around the weights from the cart to the hanger one at a time. We found that as the force of the cart increased the acceleration increased.
6) In this unit we also learned about skydiving. Skydiving is like free fall, but weight matters, and air resistance is a large factor. As a person jumps out of a plane, they are pulled down by the F-weight (which is found using the formula w=mg) and they accelerate towards earth. F-air (the force of air resistance) acts in the opposite direction and increases as the person gains speed. This is because acceleration is directly proportional to force. In this situation the force is air resistance and acceleration is how fast their velocity is increasing. F-air increases until it becomes equal to the constant F-weight. This is called terminal velocity. In terminal velocity, the net force is 0 which means the person is no longer accelerating (although they continue to move downward) and is in equilibrium. Once the person opens their parachute, the F-air becomes much greater than before, the person is no longer in equilibrium, and the speed of the person slows down. Because the speed is decreasing, air resistance decreases also because they are directly proportional. These two factors continue to decrease until F-air is equal to F-weight again. This is the second terminal velocity. In the second terminal velocity, the speed is slower than in the first and the air resistance is larger than in the first. The person continues to fall toward the ground at this much slower speed and can safely land on the ground. VIDEO HERE!!
F-air and F-weight are important parts of falling things besides sky diving.
If you dropped a crumpled piece of paper and a flat piece of paper, although they are the same weight, the crumpled paper would land first. This is because of surface area. The crumpled piece of paper has a small surface area, so it would have to accelerate longer in order to reach terminal velocity and get its F-air to equal its F-weight. The flat piece of paper has a large surface area, so it would not have to accelerate long to reach terminal velocity and have an equal F-air and F-weight.
If you dropped a lead ball and a ping pong ball, although they have the same surface area, they would land at different times. This is because they have different weights. The lead ball has to accelerate for much longer in order to reach terminal velocity and get equal F-weight and F-air and the ping pong ball doesn't have to accelerate long to reach terminal velocity with equal F-weight and F-air.
It was interesting to learn how physics plays a role in a wide range of things as simple as tossing a ball into the air, as common as playing sports, and as exiting as skydiving.
F-air and F-weight are important parts of falling things besides sky diving.
If you dropped a crumpled piece of paper and a flat piece of paper, although they are the same weight, the crumpled paper would land first. This is because of surface area. The crumpled piece of paper has a small surface area, so it would have to accelerate longer in order to reach terminal velocity and get its F-air to equal its F-weight. The flat piece of paper has a large surface area, so it would not have to accelerate long to reach terminal velocity and have an equal F-air and F-weight.
If you dropped a lead ball and a ping pong ball, although they have the same surface area, they would land at different times. This is because they have different weights. The lead ball has to accelerate for much longer in order to reach terminal velocity and get equal F-weight and F-air and the ping pong ball doesn't have to accelerate long to reach terminal velocity with equal F-weight and F-air.
It was interesting to learn how physics plays a role in a wide range of things as simple as tossing a ball into the air, as common as playing sports, and as exiting as skydiving.
