What is the Y-Component of a Baseball's Acceleration at Half Its Terminal Speed?

[pedro_infante]

Homework Statement

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

Homework Equations

F=ma
mg+bv^2=ma
mg-Bv^2=ma

The Attempt at a Solution

been trying to start it but I just don't see how.

 

[Soveraign]

What is the acceleration "a" when the object is at terminal velocity?

Using your equations, what is the terminal velocity in terms of m, g, and b?

Now, what is "a" when you cut "v" in half?

 

[tomcue]

I would approach this problem by first understanding the concept of terminal speed. Terminal speed is the maximum speed that an object can reach when falling due to the balance between the gravitational force and the drag force. In this case, the drag force is proportional to the square of the velocity, which means that as the velocity increases, the drag force also increases.

When the ball is thrown straight up, it will initially accelerate upwards due to the force of gravity. However, as it reaches higher speeds, the drag force will also increase and eventually balance out the gravitational force, resulting in a constant terminal speed.

Now, let's consider the y-component of the ball's acceleration when its speed is half its terminal speed. At this point, the drag force is only half of what it would be at the terminal speed. Therefore, the net force acting on the ball would be the difference between the gravitational force and half of the drag force. This can be represented as:

Fnet = mg - (1/2)bv^2

Using the equation F=ma, we can rearrange to solve for the acceleration:

a = (mg - (1/2)bv^2)/m

Substituting in the value of g, we get:

a = (9.8 - (1/2)bv^2)/m

Therefore, the y-component of the ball's acceleration when its speed is half its terminal speed can be calculated using this equation.

As for the ball moving back down, the y-component of the acceleration would be the same, but with a negative sign. This is because the ball is now moving in the opposite direction and the acceleration would be towards the ground.

In conclusion, the y-component of the ball's acceleration when its speed is half its terminal speed and it is moving up is given by (9.8 - (1/2)bv^2)/m and when it is moving back down, it is given by -(9.8 - (1/2)bv^2)/m.