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?

a. When it is moved up a = -5g / 4

b. When it is moved back down a  = -3g / 4

Here

ma = mg - kv ^ 2

Where,

k: proportionality constant.

For the terminal velocity, we have:

a = 0

Now Clearing the value of v:

v = root (mg / k)

v ^ 2 = mg / k

The ball's speed should be half its terminal speed.

v = root (mg / 4k)

v ^ 2 = mg / 4k

a.

going up

ma = -mg - kv ^ 2

a = -g -gk / 4k

a = -g - g / 4

a = -5g / 4

b.

Going down

ma = -mg + kv ^ 2

a = -g + gk / 4k

a = -g + g / 4

a = -3g / 4

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