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10 August, 14:46

You throw a baseball straight upward. the drag force is proportional to v2. in terms of g, what is the y-component of the ball's acceleration when the ball's speed is half its terminal speed and (a) it is moving up? (b) it is moving back down?

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  1. 10 August, 15:05
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    The first thing to do is a free-body diagram in a vertical direction.

    We have then:

    ma = mg - kv ^ 2

    Where,

    k: proportionality constant.

    For the terminal velocity, we have:

    a = 0

    Clearing the value of v:

    v = root (mg / k)

    v ^ 2 = mg / k

    The ball's speed is half its terminal speed:

    v = root (mg / 4k)

    v ^ 2 = mg / 4k

    going up

    ma = - mg - kv ^ 2

    a = - g - gk / 4k

    a = - g - g / 4

    a = - 5g / 4

    a = - 5g / 4

    Going down

    ma = - mg + kv ^ 2

    a = - g + gk / 4k

    a = - g + g / 4

    a = - 3g / 4

    a = - 3g / 4
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