Finding final velocity with variable acceleration due to air resistance

[T. Haverford]

Homework Statement

Hello! This is my first time posting here but I have been viewing these forums for help in my physics class for a while. Lately I have been stuck on one single problem that I haven’t been able to find any help with anywhere online. Any who, here is the problem: a cat jumps out of a 10m tall window with all its velocity in the X direction. He has a mass of 8.17kg, a drag coefficient of .8, and a cross sectional area of 600cm^2. What is his final velocity if you do NOT ignore air resistance?

Homework Equations

I determined, knowing the density of air to be 1.2 @ 20 C, that the cats terminal velocity would be around 52m/s. From here I solved for a drag or wind resistance constant, b, by taking the mass times gravity divided by terminal velocity (b=1.5). I also know that velocity at any time can be determined with the equation V=(mg)/b{1-e^[(-bt)/m]} where m is mass, g is gravity, b is the constant from above, and t is time. The problem is that I cannot find time with the variable acceleration due to gravity. My professor suggested using the concept of jerk but I haven’t been able to find any formulas that can help me out.

The Attempt at a Solution

(attempt included in relevant equations)

Thanks in advance for any help!

 

[TSny]

Hello on your first post to PF!

Maybe you could try using your velocity as a function of time to derive an expression for the position as a function of time. Since you know the final position of the cat, you might be able to determine the time the cat reaches that position.

 

[Chestermiller]

Welcome to PF.

You need to start over. The first thing you need to do is focus on the drag force. The drag force has components in both the horizontal and vertical directions. The spatial direction of the drag force will be opposite to the spatial direction of the velocity vector. First express the magnitude of the drag force in terms of the magnitude of the velocity vector, which is the square root of the sum of the squares of the horizontal and vertical components. Then calculate the unit vector in the direction of the drag force. Multiply the unit vector by the magnitude of the drag force to express the force as a vector. Now you are finally ready to start doing your force balances in the horizontal and vertical directions. This should lead to two coupled differential equations for the derivatives of the horizontal and vertical components of velocity with respect to time.

 

[TSny]

Ah, yes. I missed that the cat jumps horizontally! Yikes. Good luck.

 

[Nickie]

Hello! It's great to see you utilizing online resources for help with your physics class. Let's take a look at the problem you are stuck on.

Firstly, it's important to note that air resistance, or drag, is a force that acts in the opposite direction of motion and depends on the velocity of the object. This means that the acceleration due to air resistance is not a constant, but rather a variable that changes as the object's velocity changes. In this case, the cat's velocity will decrease as it falls due to the increasing air resistance.

To solve for the cat's final velocity, we can use the equation you provided: V=(mg)/b{1-e^[(-bt)/m]}. However, instead of trying to find the time, we can use the concept of terminal velocity. Terminal velocity is the maximum velocity an object can reach when the drag force equals the force of gravity.

We know that at terminal velocity, the cat's acceleration due to air resistance is 0. This means that we can set the equation equal to 0 and solve for velocity. This will give us the cat's final velocity when air resistance is taken into account.

It's also important to note that the cat's initial velocity is in the x-direction, so we can use the x-component of the equation to solve for the final velocity in that direction.

I hope this helps! Let me know if you have any further questions or need clarification. Keep up the good work in your physics class.