So idk if this makes sense so I’m going to use numbers. If I take something, and shoot it let’s say a bullet. Into the air. The first second it will be slowing down at 9.81m/s if I’m not mistaking because gravity is pulling it down that fast. At the 2nd second. Will gravity be pulling it down at 9.81m/s still or will it go to 19.62m/s? Like will the speed be stacking essentially for how fast it slows down?
In: 8
No, the object’s speed will slow down but the acceleration it is receiving, due to gravity, is a constant 9.81 m/s^2. I think you are mixing speed and acceleration.
It will slow down at the same rate every second.
Technically, yes. The further you are away from an object the less gravity has an effect on it but given the small distance you’re throwing it up and the size of the Earth the difference is nearly impossible to measure.
The acceleration towards the center of mass (down) will remain constant. Its the velocity that will increase over time as the object is unsupported.
So it will always be accelerating at 9.8 m/s^2.
So if you throw the object up at 9.81 m/s (assuming zero air resistance) after 1 second the object will be stationary in the air. It will then accelerate at the same rate downward, reaching the same height you threw it from after one more second (2 seconds total). Shortly after it will hit the ground at your feet. If you are on a cliff it will continue falling. After 2 seconds of falling it will be moving at 19.62 m/s. But it will still only be accelerating at 9.81 m/s^2.
Edit to add: You are confusing velocity and acceleration. Velocity is how fast something is changing position over time. Acceleration is how much it is changing velocity over time.
>So idk if this makes sense so I’m going to use numbers. If I take something, and shoot it let’s say a bullet. Into the air. The first second it will be slowing down at 9.81m/s if I’m not mistaking because gravity is pulling it down that fast. At the 2nd second. Will gravity be pulling it down at 9.81m/s still or will it go to 19.62m/s? Like will the speed be stacking essentially for how fast it slows down?
You’re dealing in accelerations here, not in velocities. If you throw an object upward it will be slowed down (decelerated) at a rate of 9.81 m/s^2 (meters per second **per second**), meaning it will lose 9.81 m/s of velocity every second.
So if you accelerate the object upwards so that it has a velocity of 9.81m/s when it leaves your hand, it will have a velocity of 0 after one second (at its maximum height) and will hit your hand at -9.81m/s velocity after two seconds.
on the way up, every second, the speed of the bullet will be 9.8 m/s slower. until it stops. air resistance will also slow it down quite a bit too.
then it starts falling, and every second, it will move 9.8 m/s faster (downward). but again, air resistance will limit how fast it can go.
Let’s do some rounding and approximation so we have nicer easier numbers. We will also ignore air resistance, gravity changing with distance and all that stuff. Basically let’s ignore everything and assume a constant pull from earth’s gravitaty.
Earth accelerates objects downwards at about 10 m/s per second. The velocity decreases by 10 meter/second during each second. (that’s exactly what m/s^2 means)
So if you shoot upwards with a starting velocity of 500m/s, it will slow down to 490 m/s after one second. After 2 seconds it will have a speed of 480 m/s, after one more second it will be down to 470 m/s, the next second at 460 m/s etc.. Eventually after about 50 seconds it will reach it’s peak and stand still for a brief moment. During the next second it will start falling down with a speed of 10m/s. This speed will increase to 20m/s during the next second and so on. Until the bullet hits the ground with exactly 500m/s. That’s why it’s so incredibly dangerous to shoot straight up with a gun.
Speed is distance per time. Acceleration is speed per time, or distance per time per time. If speed increases at a constant rate, acceleration is constant.
If you drop something off a building, it starts off at rest. It is immediately and constantly accelerated by Earth’s gravity at 9.81 m/s^(2) (approximately), meaning its velocity will increase by 9.81 m/s every second. After 1 second, it will be falling at 9.81 m/s. After 2 seconds, it will be falling at 19.62 m/s. After 3 seconds, it will be falling at 29.43 m/s. It will keep accelerating like this until it is stopped by some other force.
An easy way to think about it is to imagine the relationship between distance and speed instead of speed and acceleration. They’re the same thing. Imagine a ball rolling along a surface. If there are no forces, it’ll keep going at the same speed forever, but the distance it covers will increase at a constant rate. At the beginning, it will have covered 0 metres. After 1 second, it will have covered, for example, 10 metres. After 2 seconds, it will have covered 20 metres. Its speed is constant, which causes its distance to increase constantly.
Now put an extra “per time” on each of the variables like how I explained up at the top. The falling object’s acceleration (speed per time) is constant, which causes its speed (distance per time) to increase constantly. It’s the same concept, but it seems more complex because we’re dealing either constantly changing values that are hard to measure.
[Obligatory Vsauce video.](https://youtu.be/U6VBV4QUMu0)
So idk if this makes sense so I’m going to use numbers. If I take something, and shoot it let’s say a bullet. Into the air. The first second it will be slowing down at 9.81m/s if I’m not mistaking because gravity is pulling it down that fast. At the 2nd second. Will gravity be pulling it down at 9.81m/s still or will it go to 19.62m/s? Like will the speed be stacking essentially for how fast it slows down?
In: 8
No, the object’s speed will slow down but the acceleration it is receiving, due to gravity, is a constant 9.81 m/s^2. I think you are mixing speed and acceleration.
It will slow down at the same rate every second.
Technically, yes. The further you are away from an object the less gravity has an effect on it but given the small distance you’re throwing it up and the size of the Earth the difference is nearly impossible to measure.
The acceleration towards the center of mass (down) will remain constant. Its the velocity that will increase over time as the object is unsupported.
So it will always be accelerating at 9.8 m/s^2.
So if you throw the object up at 9.81 m/s (assuming zero air resistance) after 1 second the object will be stationary in the air. It will then accelerate at the same rate downward, reaching the same height you threw it from after one more second (2 seconds total). Shortly after it will hit the ground at your feet. If you are on a cliff it will continue falling. After 2 seconds of falling it will be moving at 19.62 m/s. But it will still only be accelerating at 9.81 m/s^2.
Edit to add: You are confusing velocity and acceleration. Velocity is how fast something is changing position over time. Acceleration is how much it is changing velocity over time.
>So idk if this makes sense so I’m going to use numbers. If I take something, and shoot it let’s say a bullet. Into the air. The first second it will be slowing down at 9.81m/s if I’m not mistaking because gravity is pulling it down that fast. At the 2nd second. Will gravity be pulling it down at 9.81m/s still or will it go to 19.62m/s? Like will the speed be stacking essentially for how fast it slows down?
You’re dealing in accelerations here, not in velocities. If you throw an object upward it will be slowed down (decelerated) at a rate of 9.81 m/s^2 (meters per second **per second**), meaning it will lose 9.81 m/s of velocity every second.
So if you accelerate the object upwards so that it has a velocity of 9.81m/s when it leaves your hand, it will have a velocity of 0 after one second (at its maximum height) and will hit your hand at -9.81m/s velocity after two seconds.
on the way up, every second, the speed of the bullet will be 9.8 m/s slower. until it stops. air resistance will also slow it down quite a bit too.
then it starts falling, and every second, it will move 9.8 m/s faster (downward). but again, air resistance will limit how fast it can go.
Let’s do some rounding and approximation so we have nicer easier numbers. We will also ignore air resistance, gravity changing with distance and all that stuff. Basically let’s ignore everything and assume a constant pull from earth’s gravitaty.
Earth accelerates objects downwards at about 10 m/s per second. The velocity decreases by 10 meter/second during each second. (that’s exactly what m/s^2 means)
So if you shoot upwards with a starting velocity of 500m/s, it will slow down to 490 m/s after one second. After 2 seconds it will have a speed of 480 m/s, after one more second it will be down to 470 m/s, the next second at 460 m/s etc.. Eventually after about 50 seconds it will reach it’s peak and stand still for a brief moment. During the next second it will start falling down with a speed of 10m/s. This speed will increase to 20m/s during the next second and so on. Until the bullet hits the ground with exactly 500m/s. That’s why it’s so incredibly dangerous to shoot straight up with a gun.
Speed is distance per time. Acceleration is speed per time, or distance per time per time. If speed increases at a constant rate, acceleration is constant.
If you drop something off a building, it starts off at rest. It is immediately and constantly accelerated by Earth’s gravity at 9.81 m/s^(2) (approximately), meaning its velocity will increase by 9.81 m/s every second. After 1 second, it will be falling at 9.81 m/s. After 2 seconds, it will be falling at 19.62 m/s. After 3 seconds, it will be falling at 29.43 m/s. It will keep accelerating like this until it is stopped by some other force.
An easy way to think about it is to imagine the relationship between distance and speed instead of speed and acceleration. They’re the same thing. Imagine a ball rolling along a surface. If there are no forces, it’ll keep going at the same speed forever, but the distance it covers will increase at a constant rate. At the beginning, it will have covered 0 metres. After 1 second, it will have covered, for example, 10 metres. After 2 seconds, it will have covered 20 metres. Its speed is constant, which causes its distance to increase constantly.
Now put an extra “per time” on each of the variables like how I explained up at the top. The falling object’s acceleration (speed per time) is constant, which causes its speed (distance per time) to increase constantly. It’s the same concept, but it seems more complex because we’re dealing either constantly changing values that are hard to measure.
[Obligatory Vsauce video.](https://youtu.be/U6VBV4QUMu0)
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