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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?

Respuesta :

carlosego
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

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

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

Calculation of y-component of the ball's acceleration:

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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