Calculating Y-Component of Acceleration for a Thrown Ball

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SUMMARY

The discussion focuses on calculating the y-component of acceleration for a baseball thrown vertically, considering drag force proportional to the square of velocity. The terminal velocity is defined as \( v_{terminal} = \sqrt{\frac{mg}{D}} \). When the ball's speed is half its terminal speed while moving upwards, the y-component of acceleration can be expressed as \( a_y = g - \frac{D}{m} \left(\frac{1}{2}v_{terminal}\right)^2 \). Conversely, when descending at the same speed, the y-component of acceleration is \( a_y = -g + \frac{D}{m} \left(\frac{1}{2}v_{terminal}\right)^2 \).

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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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