You throw a baseball straight upward. The drag force is proportional to v 2 v^2 v 2 . In terms of g g g , what is the y y y -component of the ball's acceleration when the ball's speed is half its terminal speed and it is moving back down?

Welcome back everybody. We have a spherical stone that is launched vertically upward. And after some time what goes up must come down now, when it is going down right, we are going to have some sort of vertical acceleration acting downward and we need to find what this is. Well, when the stone is falling there are a couple of forces acting on it. We have a downwards force and an upwards force downwards force is of course the force due to gravity, but the upwards force is a drag force. Now this drag force is proportional to the terminal velocity squared by some proportionality constant, which I'm just going to call C. Now we want to find this vertical acceleration at a speed that is half of the terminal velocity. So what is our A Y. At this speed? Right here, let's use Newton's second law here. Newton's second law states that the sum of all forces in a certain direction, we're gonna look at the Y direction in this case is equal to mass times some acceleration. And this is the acceleration that we are looking for. Now, plugging in our values, we have our drag force minus are forced due to gravity equal to our mass times are vertical acceleration, but it's going to be negative since it is in the negative y direction. Now dividing both sides by negative M. We get that our vertical acceleration is equal to our force due to gravity minus our drag force all over our mass. But what is our drag force? Well, we're told that at terminal velocity a terminal velocity, our drag force is equal to our force to gravity. But we're not looking at terminal velocity. We're looking at this speed right here. Right? So just as a reminder, our drag force which is equal to our force due to gravity at terminal velocity is proportional to our terminal velocity squared. But let's sub in this velocity right here, shall we? So our drag force equal to our proportionality constant times. Are terminal velocity over two squared equal to 0.25 times C squared. Well at terminal velocity C v t squared is just equal to MG. So this is just equal to 0.25 MG. Now let's plug this into our formula four A. Y. We have that A. Y. Is equal to MG minus our drag force equal to 0.25 MG. All divided by our mass Yielding us 0.75GS corresponding to our answer choice of E. Thank you all so much for watching. Hope this video helped. We will see you all in the next one

7. Friction, Inclines, Systems

Kinetic Friction

Physics

7. Friction, Inclines, Systems

Kinetic Friction

3:36 minutes

Problem 40b

Textbook Question

You throw a baseball straight upward. The drag force is proportional to $v^2$ . In terms of $g$ , what is the $y$ -component of the ball's acceleration when the ball's speed is half its terminal speed and it is moving back down?

Verified step by step guidance

Step 1: Understand the forces acting on the baseball. When the ball is moving upward or downward, two forces act on it: gravity (mg, acting downward) and the drag force (proportional to υ², acting opposite to the direction of motion). Terminal velocity occurs when the drag force equals the gravitational force, meaning the net acceleration is zero.

Step 2: Express the drag force mathematically. The drag force can be written as F_drag = kυ², where k is a proportionality constant and υ is the velocity of the ball. At terminal velocity (υ_terminal), the drag force equals the gravitational force: kυ_terminal² = mg.

Step 3: Relate the drag force to the current velocity. When the ball's speed is half its terminal velocity (υ = 0.5υ_terminal), the drag force becomes F_drag = k(0.5υ_terminal)² = k(υ_terminal²/4). Substitute kυ_terminal² = mg into this expression to find F_drag = mg/4.

Step 4: Determine the net force and acceleration when the ball is moving upward. The net force is the sum of the forces acting on the ball: F_net = -mg (gravity, downward) - F_drag (drag, downward). Substituting F_drag = mg/4, the net force becomes F_net = -mg - mg/4 = -(5/4)mg. Use Newton's second law (F_net = ma) to find the acceleration: a = F_net/m = -(5/4)g.

Step 5: Determine the net force and acceleration when the ball is moving downward. In this case, gravity acts downward (positive direction), and the drag force acts upward (negative direction). The net force is F_net = mg (gravity, downward) - F_drag (drag, upward). Substituting F_drag = mg/4, the net force becomes F_net = mg - mg/4 = (3/4)mg. Use Newton's second law to find the acceleration: a = F_net/m = (3/4)g.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Terminal Velocity

Terminal velocity is the constant speed an object reaches when the drag force acting on it equals the gravitational force, resulting in zero net acceleration. For a baseball thrown upward, this speed is influenced by its mass, shape, and the density of the air. Understanding terminal velocity is crucial for analyzing the motion of the baseball as it ascends and descends.

Drag Force

The drag force is a resistive force that opposes the motion of an object through a fluid, such as air. It is often modeled as being proportional to the square of the object's velocity (υ²). This concept is essential for determining how the baseball's speed affects its acceleration, especially when considering forces acting on it during its flight.

Acceleration due to Gravity

Acceleration due to gravity (g) is the rate at which an object accelerates towards the Earth when in free fall, approximately 9.81 m/s². This acceleration acts downward and is a key factor in the motion of the baseball. When analyzing the ball's motion, both the gravitational force and the drag force must be considered to determine the net acceleration at various speeds.