Calculating Y-Component of Acceleration for a Thrown Ball

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To calculate the y-component of acceleration for a baseball thrown straight up, consider the effects of gravity and drag force, which is proportional to the square of the velocity. When the ball's speed is half its terminal speed while moving upward, the net force can be expressed as the difference between gravitational force and drag force. Conversely, when the ball is moving downward at half its terminal speed, the drag force acts in the opposite direction to gravity. The equations of motion and the relationship between forces must be applied to derive the acceleration in both scenarios. Understanding these dynamics is crucial for solving the problem effectively.
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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? 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 back down?

Homework Equations



v terminal = \sqrt{mg/D}

The Attempt at a Solution


I honestly have no idea how to attempt this problem
 
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The drag force is of the form:
F_d = -kv^2

Or to use your notation:
F_d = -Dv^2

The terminal velocity is achieved when the ball starts falling at a constant velocity. At that point, the following holds true:
kv^2=mg
 
clarineterr said:
I honestly have no idea how to attempt this problem

Hi clarineterr! :wink:

Start "acceleration = dv/dt = … " :smile:
 
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