Quadratic Drag Force: Solving Differently

In summary, the first question involves solving for speed as a function of position during ascent, using the equation of motion $$m\ddot y = -mkv^2-mg$$. This matches the answer to the second question, which involves solving for speed as a function of time using $$m\dot v = mg - cv^2$$. However, the first question requires going further and finding speed as a function of position, which is made easier by using the equation $$\frac{dv}{dt}=\frac{dv}{dy}\frac{dy}{dt}$$. This is because it is generally difficult to solve drag equations for position as a function of time, but solving for speed as a function of position is more
  • #1
tryingtolearn1
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Homework Statement
There are two questions in my book that are almost the same but have two different answers.

The first question:
A particle is projected vertically upward in a constant gravitational field with an initial speed of ##v0##. Show that if there is a retarding force proportional to the square of the speed, the speed of the particle when it returns to its initial position is where $$\frac{v_ov_T}{\sqrt{v_o^2+v_T^2}}$$ ##v_T## is the terminal speed.

Second question:
A baseball is thrown vertically up with speed v o and is subject to a quadratic drag with magnitude ##f (v) = cv^2## . Write down the equation of motion for the upward journey (measuring ##y## vertically up).
Relevant Equations
##F=ma##
But the answers to these two questions confuses me.

For the first question the answer goes like: $$m\ddot y = -mkv^2-mg$$ $$\therefore \ddot y = v\frac{dv}{dy}$$ $$\therefore v\frac{dv}{dy}= -kv^2-g$$

but for the answer to the second question we have: $$m\dot v = mg - cv^2.$$

Both questions are very similar but with different answers. Why is the first question solving it using ##\frac{dv}{dy}\frac{dy}{dt}## instead of just ##\dot v##?
 
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  • #2
The first equation for the first question is the equation of motion during ascent, and it matches the answer to the second question. But the first question requires you to go further.
It is not in general possible, or at least not easy, to solve drag equations to get position as a function of time. But in the first problem you do not need speed or position as a function of time, but speed as a function of position. Using ##\frac{dv}{dt}=\frac{dv}{dy}\frac{dy}{dt}## makes it a lot easier.
 
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Related to Quadratic Drag Force: Solving Differently

1. What is quadratic drag force?

Quadratic drag force is a type of drag force that is proportional to the square of the velocity of an object moving through a fluid, such as air or water. It is caused by the resistance of the fluid against the object's motion.

2. How is quadratic drag force different from other types of drag force?

Unlike other types of drag force, such as linear or viscous drag, quadratic drag force increases exponentially as the velocity of the object increases. This means that as the object moves faster, the drag force becomes significantly stronger.

3. How is quadratic drag force calculated?

The formula for calculating quadratic drag force is Fd = ½ * ρ * v^2 * Cd * A, where Fd is the drag force, ρ is the density of the fluid, v is the velocity of the object, Cd is the drag coefficient, and A is the cross-sectional area of the object.

4. Can quadratic drag force be solved for differently?

Yes, there are different methods for solving quadratic drag force, such as using the drag equation or using numerical methods like Euler's method. These methods may be more accurate for certain scenarios, such as when the velocity of the object is changing over time.

5. What are some real-life examples of quadratic drag force?

Quadratic drag force can be observed in various real-life situations, such as the drag experienced by a car moving at high speeds, the resistance faced by a swimmer in water, or the force acting on a rocket as it moves through the atmosphere. It is also important to consider in the design of objects like airplanes and sports equipment.

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