I recently got access to Bing and asked it about the bullet in temporary gravity of varying duration. It does quite a bit better than GPT-3 though it's very verbose. It does do a search during it's answer but only to find the typical initial velocity of a bullet. It makes an error regarding the final velocity of the bullet after three seconds, but correctly determines that the bullet will go up forever if gravity lasts three seconds but will fall back to Earth if it lasts five minutes. Bold is me, everything else is Bing.

Okay, I’ve cleared the slate for a fresh start. What can I help you explore now?

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Let's think of a physics problem, Neglect only air resistance. Suppose I shoot a bullet straight into the sky. Imagine that the laws of physics work a little differently: during the first three seconds after I shoot the bullet, gravity is present. After three seconds, gravity stops applying forever. Where does the bullet go?

Searching for: bullet initial velocity

Searching for: bullet initial velocity

Generating answers for you…

Generating answers for you…

That’s an interesting physics problem. To solve it, we need to know some information about the bullet and its motion. For example, what is the initial velocity of the bullet when you shoot it? According to ¹, a typical bullet can have an initial velocity of 1200–1500 m/s. What kind of bullet are you using?

Received message. That's an interesting physics problem. To solve it, we need to know some information about the bullet and its motion. For example, what is the initial velocity of the bullet when you shoot it? According to , a typical bullet can have an initial velocity of **1200–1500 m/s**. What kind of bullet are you using?

Learn more:1. hypertextbook.com2. gundigest.com3. sage-answer.com4. sarthaks.com5. wikihow.com

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Let's suppose the initial velocity of the bullet is 1500 m/s

Okay, let’s assume that. Then we can use the equations of motion to find out where the bullet goes. Since you are shooting the bullet straight into the sky, we can ignore the horizontal component of its motion and focus on the vertical component. The equation for the vertical displacement of the bullet is:

y = y0 + v0t - 0.5gt^2

where y is the final height of the bullet, y0 is the initial height of the bullet (which we can assume to be zero), v0 is the initial velocity of the bullet (which you said is 1500 m/s), t is the time elapsed and g is the acceleration due to gravity (which we can assume to be 9.8 m/s^2).

We want to find out what happens to the bullet after three seconds when gravity stops applying forever. So we need to find out how high the bullet has reached at that point. To do that, we plug in t = 3 into the equation and get:

y = 0 + 1500 * 3 - 0.5 * 9.8 * 3^2

Simplifying this gives us:

y = 4500 - 44.1 * 9

or

y = 4096.1 m

So after three seconds, when gravity stops applying forever, the bullet has reached a height of 4096.1 meters above ground. What do you think happens next?

what do you think happens?

Well, since gravity stops applying forever after three seconds, there is nothing to pull the bullet down or slow it down. So the bullet will continue to move with the same velocity and direction that it had at that moment. That means the bullet will keep going up forever with a constant speed of 1500 m/s. It will never stop or fall down.

Does that make sense?

Close, it will probably be going a bit slower than that because gravity will have slowed it down in the first three seconds, but I agree it will keep going up forever.

You are right, the bullet will have a slightly lower speed than 1500 m/s after three seconds because of gravity. To find out the exact speed, we can use another equation of motion:

v = v0 - gt

where v is the final velocity of the bullet, v0 is the initial velocity of the bullet (which we said is 1500 m/s), t is the time elapsed and g is the acceleration due to gravity (which we said is 9.8 m/s^2).

Plugging in t = 3 into this equation gives us:

v = 1500 - 9.8 * 3

or

v = 1470.6 m/s

So after three seconds, when gravity stops applying forever, the bullet has a speed of 1470.6 m/s and it will keep that speed forever as it goes up.

Do you have any other physics problems you want to explore?

Great! What would happen to the bullet if, instead, gravity lasted for five minutes before disappearing?

That’s another interesting scenario. If gravity lasted for five minutes before disappearing, then the bullet would behave differently. First of all, we need to find out if the bullet would still be in the air after five minutes or if it would have fallen back to the ground by then. To do that, we need to find out how long it takes for the bullet to reach its maximum height and start falling down. We can use the same equation of motion as before:

v = v0 - gt

but this time we set v to zero because that’s when the bullet stops going up and starts going down. Solving for t gives us:

t = v0 / g

Plugging in v0 = 1500 m/s and g = 9.8 m/s^2 gives us:

t = 1500 / 9.8

or

t = 153.1 s

So it takes 153.1 seconds or about 2.6 minutes for the bullet to reach its maximum height and start falling down.

Now we need to compare this time with the five minutes that gravity lasts before disappearing. Since five minutes is longer than 2.6 minutes, that means gravity will still be present when the bullet is falling down.

To find out what happens next, we need to use another equation of motion:

y = y0 + v0t - 0.5gt^2

but this time we set y to zero because that’s when the bullet hits the ground again. Solving for t gives us a quadratic equation:

-0.5gt^2 + v0t - y0 = 0

We can use the quadratic formula to find the solutions for t. The formula is:

t = (-b ± √(b^2 - 4ac)) / (2a)

where a = -0.5g, b = v0 and c = -y0.

Plugging in our values gives us:

t = (-1500 ± √(1500^2 - 4 * (-0.5 * 9.8) * 0)) / (2 * (-0.5 * 9.8))

Simplifying this gives us:

t = (-1500 ± √(2250004)) / (-9.8)

or

t ≈ -306 s or +306 s

We can ignore the negative solution because it doesn’t make sense in our context (it means going back in time). So the positive solution tells us that it takes 306 seconds or about 5.1 minutes for the bullet to hit the ground again.

Now we can compare this time with the five minutes that gravity lasts before disappearing. Since five minutes is shorter than 5.1 minutes, that means gravity will disappear before the bullet hits the ground again.

To find out what happens next, we need to find out how high and fast the bullet is at exactly five minutes after being shot into the sky.

To do that, we can use our equations of motion again but plug in t=300 (which is five minutes in seconds).

For height:

y= y_ {initial}+v_ {initial} t- \frac{1}{2} g t^{2}

For height:

y= y_ {initial}+v_ {initial} t- \frac{1}{2} g t^{2}

Plugging in our values gives us:

y= 15000*300- \frac{1}{2}*9 .8*300^{2}

Simplifying this gives us:

y=450000-441000

or

y=9000 m

So after five minutes, when gravity disappears forever, the bullet has a height of 9000 meters above ground.

For speed:

v=v_ {initial}-g t

Plugging in our values gives us:

v=15000-9 .8*300

Simplifying this gives us:

v=-14400 m/s

So after five minutes, when gravity disappears forever, the bullet has a speed of -14400 m/s which means it is still falling down with a very high speed.

What do you think happens next?

what do you think?

Well, since gravity disappears forever after five minutes, there is nothing to pull the bullet down or slow it down. So the bullet will continue to move with the same velocity and direction that it had at that moment. That means the bullet will keep falling down forever with a constant speed of -14400 m/s. It will never stop or go up.

Does that make sense?