What is the terminal velocity?

In summary, the conversation discusses a projectile being ejected into an experimental fluid with an initial speed and angle. The drag force on the projectile results in an acceleration term and the conversation explores determining the x- and y-components of velocity and displacement as functions of time. The terminal velocity is also mentioned, including the effects of gravitational acceleration. The conversation ends with a statement about not being sure of the correct approach to solving the problem.
  • #1
Warr
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A projectile is ejected into an experimental fluid at time t = 0. The initial speed is v0 and the angle to the horizontal is theta. The drag in the projectile results in an acceleration term a_d=-kv, where k is a constant and v is the velocity of the projectile. Determine the x- and y-components of both the velocity and displacement as functions of time. What is the terminal velocity? Include the effects of gravitational acceleration.

here's a diagram:

attachment.php?attachmentid=4891&stc=1.jpg


I don't really know what I'm doing here.

What I tried was splitting it into components and solving the dif eqs. I didn't know what 'constants' of integration would do so I just left them out.

a_x = -kv_x
v_x = e^(-kt)

a_y = -kv_y - g
v_y = -(gt)e^(-k*t)

let me just say I seriously do not know what I'm doing. Thus far we hadn't had any dif. eqs in the examples in class. No drag forces with gravity or anything. Btw this has to be solved with dynamics not statics I believe. I'm probably going in the completely wrong direction..
 

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  • #2
Warr said:
A projectile is ejected into an experimental fluid at time t = 0. The initial speed is v0 and the angle to the horizontal is theta. The drag in the projectile results in an acceleration term a_d=-kv, where k is a constant and v is the velocity of the projectile. Determine the x- and y-components of both the velocity and displacement as functions of time. What is the terminal velocity? Include the effects of gravitational acceleration.

I don't really know what I'm doing here.

What I tried was splitting it into components and solving the dif eqs. I didn't know what 'constants' of integration would do so I just left them out.

a_x = -kv_x
v_x = e^(-kt)

a_y = -kv_y - g
v_y = -(gt)e^(-k*t)

let me just say I seriously do not know what I'm doing. Thus far we hadn't had any dif. eqs in the examples in class. No drag forces with gravity or anything. Btw this has to be solved with dynamics not statics I believe. I'm probably going in the completely wrong direction..
Resolving into components:

[tex]k\dot s_x + \ddot s_x = 0[/tex] the solution of which is:

[tex]v_x = v_{x0}e^{-kt}[/tex]

and:

[tex]k\dot s_y + \ddot s_y = g[/tex]

You can work out that solution, but you can also see that when kt is large, [itex]v_x = 0[/tex] and when kvy = g or vy = g/k there is zero acceleration and the object reaches terminal velocity.

AM
 
  • #3


Dear student,

Thank you for your question. Terminal velocity is defined as the maximum velocity that an object reaches when falling through a fluid due to gravity and air resistance (or drag force) balancing each other out. In other words, it is the point at which the acceleration of the object becomes zero and the object falls at a constant velocity.

In this scenario, the terminal velocity can be determined by setting the acceleration equal to zero and solving for the velocity. Using the given information, we can write the equation for the acceleration of the projectile as follows:

a = a_d + a_g

Where a_d is the drag acceleration and a_g is the gravitational acceleration. Substituting the given values, we have:

0 = -kv - g

Solving for v, we get:

v = -g/k

This is the terminal velocity of the projectile. It is important to note that the terminal velocity depends on the value of k, which is determined by the shape and size of the projectile, as well as the properties of the fluid it is falling through.

To determine the x- and y-components of velocity and displacement as functions of time, we can use the equations of motion:

v_x = v0cos(theta)e^(-kt)

v_y = -g/k + v0sin(theta)e^(-kt)

x = v0cos(theta)(1-e^(-kt))/k

y = (-g/k + v0sin(theta))(1-e^(-kt))/k + (gt^2)/2

Again, it is important to note that these equations will vary depending on the value of k. Additionally, the constants of integration are necessary to fully solve the equations, so it is important to include them in your calculations.

I hope this helps clarify the concept of terminal velocity and how it relates to the given scenario. Please let me know if you have any further questions. Good luck with your studies!
 

1. What is terminal velocity?

Terminal velocity is the maximum speed that an object will reach when falling through a fluid (such as air or water) due to the balance of gravity and air resistance.

2. How is terminal velocity calculated?

Terminal velocity is calculated using the formula Vt = √(2mg/ρACd), where Vt is the terminal velocity, m is the mass of the object, g is the acceleration due to gravity, ρ is the density of the fluid, A is the projected area of the object, and Cd is the drag coefficient.

3. What factors affect terminal velocity?

The factors that affect terminal velocity include the mass and shape of the object, the density and viscosity of the fluid, and the gravitational pull on the object.

4. Can an object reach a higher or lower terminal velocity?

An object can reach a higher or lower terminal velocity depending on the factors mentioned before. For example, a heavier object will have a higher terminal velocity than a lighter object, and a more aerodynamic object will have a lower terminal velocity than a less aerodynamic object.

5. Is terminal velocity the same for all objects?

No, terminal velocity will vary for different objects depending on the factors mentioned before. For example, a feather will have a lower terminal velocity than a brick due to differences in mass, shape, and air resistance.

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