How Do Drag and Gravity Affect Projectile Motion?

[Warr]

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:

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..

 

[Andrew Mason]

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:

$$k\dot s_x + \ddot s_x = 0$$ the solution of which is:

$$v_x = v_{x0}e^{-kt}$$

and:

$$k\dot s_y + \ddot s_y = g$$

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

AM

 

[quicknote]

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!