Finding gravity

[Mea quidem sententia]

I've been trying to find out how to find gravity based on a projectile being fired horizontally. I am looking for g, but not with the equation of gravitational acceleration, since that seems to be only applicable to planets and stars. I'm not dealing with these, but with a ship where the gravity is higher than 1 g. I only know it's greater than 1 g, but not by how much.

So, here's the information that I have. The projectile is traveling at 343 m/s and the range for this projectile is 10 meters. However, it only covers 4.551 meters due to the increased gravity. The distance between the person's arm firing the projectile and the ground is approximately 1.296 meters.

I have seen an equation for gravity written as g = 2(viyt - dy) / t^2. To clarify those odd looking parts, viy is the vertical initial velocity, since y is vertical on the axis. t is time. dy is a change in y divided by time squared. I can't say this is useful, however, and I'm guessing in order to figure out by how much g is increased, I'd be using kinematics.

If anyone can help me figure out how much greater this ship's gravity is to Earth's, Earth's being 1 g, I'd really appreciate it. Even more so if I can see all the steps.

[Morberticus]

I'm not sure I understand the question. You fire a projectile at 343 m/s, but it only covers 4.551 meters? I.e. It's only been in the air for 4.551/343 = 0.0132 seconds?

This means the projectile, under the force of gravity, would have travelled 1.296 metres vertically in that time.

A useful equation (which is equivalent to the one you posted) would be

distance = initial velocity * time + 0.5 * g * time^2

Initial (vertical) velocity is 0, so it reduces to

distance = 0.5 * g * time^2

g would then be 14876 m/s^2, or approximately 1,500 times greater than earth's gravity.

I am almost certain I have misinterpreted the question. What do you mean when you say the projectile has a range of 10 m under earth's gravity? If you mean it would have normally gotten as far as 10 m, but only got 4.551 m, due the increased gravity, then g would be 4.394 times stronger than earth's gravity.

[Mea quidem sententia]

Morberticus

I'm not sure I understand the question. You fire a projectile at 343 m/s, but it only covers 4.551 meters? I.e. It's only been in the air for 4.551/343 = 0.0132 seconds?

This means the projectile, under the force of gravity, would have travelled 1.296 metres vertically in that time.

A useful equation (which is equivalent to the one you posted) would be

distance = initial velocity * time + 0.5 * g * time^2

Initial (vertical) velocity is 0, so it reduces to

distance = 0.5 * g * time^2

g would then be 14876 m.s^2, or approximately 1,500 times greater than earth's gravity.

I am almost certain I have misinterpreted the question. What do you mean when you say the projectile has a range of 10 m under earth's gravity? If you mean it would have normally gotten as far as 10 m, but only got 4.551 m, due the increased gravity, then g would be 4.394 times stronger than earth's gravity.

Yeah, it only covers 4.551 meters instead of 10 meters because the gravity is greater. I just wasn't sure how much gravity.

I know of d = vit 1/2 at^2, which I also can be written as d = gt^2 / 2, and I was thinking of using this, but I only understand it in regard to the y axis, simply because it's used to find the distance of a falling object if all you know is the gravity and time.

I wanted to use the kinematic equation you used, but I wasn't sure how to do it because if I am looking for g (9.8 m/s^2), why would I use g? I'm guessing the time you used was 0.0132 s so that d = gt^2 / 2 would be written as d = (9.8 m/s^2)(0.0132 s)^2 / 2. That would be 0.000853776 m.

But yes, that's what I meant by 10 meters under Earth's gravity. How did you get 4.394 times greater than Earth's gravity?

[Mea quidem sententia]

Morberticus

It took me time, but I think I figured it out! I don't know how you did it, but what do you think? First, I divided 4.551 meters / 343 m/s and then did it again with 10 m. / 343 m/s. Then I used g = 2(viy - dy) / t^2.

4.551 m. / 343 m/s = 0.0132 s
10 m. / 343 m/s = 0.0292 s

The distance between the person's arm and the ground is 1.296 m. Since the initial velocity and time are 0, I did it the following way.

g = 2(0 - 1.296) / 0.0132^2
g = 2(-1.296) / 0.0132^2
g = -2.592 / 0.0132^2
g = -2.592 / 0.00017424
g = -14,876

That's what you got, except there wasn't a negative sign in front of it. Then I did it again, but using the time I got by dividing 10 m. / 343 m/s.

g = 2(0 - 1.296) / 0.0292^2
g = 2(-1.296) / 0.0292^2
g = -2.592 / 0.0292^2
g = -2.592 / 0.00085264
g = -3,040

After that, I did the following.

-14,876 / -3,040 = 4.89

I know you ended up with 4.39 g, but looking at someone else's calculation from September 19, 2012, he ended up with 4.82 g. So, it's not far off, but I want to make sure I did it correctly. I feel that I did, and if I did, then I am very pleased!

[Morberticus]

The problem is both scenarios have a gravitational force much much greater than Earth's, which has me confused. Is the 10 m range supposed to be the range of the weapon on earth?

Anyway, this is my computation.

For a fixed horizontal velocity v, the projectile travels a) 10 m, b) 4.551 m. Therefore, since v is the same in both cases, t a) must be 10/4.551 or 2.197 * t b) I.e. t_a = 2.197 t_b

Using the formula

falling distance s = 0.5 * g * t^2
g = 2 * s / t^2

We see that, for scenario a),

g = 2 * s / (t_a)^2 = g_a

For scenario b)

g = 2 * s/(t_b)^2 = 2*s/(2.197 t_a)^2 = 1/4.828 g_a = g_b

Therefore, we can say that g_b = 4.828 g_a

The 4.394 was me not being able to use a calculator.

Note that this result is independent of the falling distance s or initial velocity v. If scenario a) is earth, the initial velocity can't be 300+ m/s

[Mea quidem sententia]

Morberticus

The problem is both scenarios have a gravitational force much much greater than Earth's, which has me confused. Is the 10 m range supposed to be the range of the weapon on earth?

Anyway, this is my computation.

For a fixed horizontal velocity v, the projectile travels a) 10 m, b) 4.551 m. Therefore, since v is the same in both cases, t a) must be 10/4.551 or 2.197 * t b) I.e. t_a = 2.197 t_b

Using the formula

falling distance s = 0.5 * g * t^2
g = 2 * s / t^2

We see that, for scenario a),

g = 2 * s / (t_a)^2 = g_a

For scenario b)

g = 2 * s/(t_b)^2 = 2*s/(2.197 t_a)^2 = 1/4.828 g_a = g_b

Therefore, we can say that g_b = 4.828 g_a

The 4.394 was me not being able to use a calculator.

Note that this result is independent of the falling distance s or initial velocity v. If scenario a) is earth, the initial velocity can't be 300+ m/s

Yeah, 10 meters is the distance given under Earth's gravity. I went with the initial velocity being 0 m/s and the final velocity being 343 m/s.

[Morberticus]

Mea quidem sententia

Morberticus

The problem is both scenarios have a gravitational force much much greater than Earth's, which has me confused. Is the 10 m range supposed to be the range of the weapon on earth?

Anyway, this is my computation.

For a fixed horizontal velocity v, the projectile travels a) 10 m, b) 4.551 m. Therefore, since v is the same in both cases, t a) must be 10/4.551 or 2.197 * t b) I.e. t_a = 2.197 t_b

Using the formula

falling distance s = 0.5 * g * t^2
g = 2 * s / t^2

We see that, for scenario a),

g = 2 * s / (t_a)^2 = g_a

For scenario b)

g = 2 * s/(t_b)^2 = 2*s/(2.197 t_a)^2 = 1/4.828 g_a = g_b

Therefore, we can say that g_b = 4.828 g_a

The 4.394 was me not being able to use a calculator.

Note that this result is independent of the falling distance s or initial velocity v. If scenario a) is earth, the initial velocity can't be 300+ m/s

Yeah, 10 meters is the distance given under Earth's gravity. I went with the initial velocity being 0 m/s and the final velocity being 343 m/s.

(I messed up the labeling in my calculations but you get the idea).

Final velocity wouldn't be 343 m/s either. I think either the initial velocity of 343 m/s or, more likely, the 10 metre range is a mistake on the part of the person posing the question. 343 m/s is the speed of sound. Fire a bullet at that speed and it would certainly have range longer than 10 m.

[Mea quidem sententia]

Morberticus

Mea quidem sententia

Morberticus

The problem is both scenarios have a gravitational force much much greater than Earth's, which has me confused. Is the 10 m range supposed to be the range of the weapon on earth?

Anyway, this is my computation.

For a fixed horizontal velocity v, the projectile travels a) 10 m, b) 4.551 m. Therefore, since v is the same in both cases, t a) must be 10/4.551 or 2.197 * t b) I.e. t_a = 2.197 t_b

Using the formula

falling distance s = 0.5 * g * t^2
g = 2 * s / t^2

We see that, for scenario a),

g = 2 * s / (t_a)^2 = g_a

For scenario b)

g = 2 * s/(t_b)^2 = 2*s/(2.197 t_a)^2 = 1/4.828 g_a = g_b

Therefore, we can say that g_b = 4.828 g_a

The 4.394 was me not being able to use a calculator.

Note that this result is independent of the falling distance s or initial velocity v. If scenario a) is earth, the initial velocity can't be 300+ m/s

Yeah, 10 meters is the distance given under Earth's gravity. I went with the initial velocity being 0 m/s and the final velocity being 343 m/s.

(I messed up the labeling in my calculations but you get the idea).

Final velocity wouldn't be 343 m/s either. I think either the initial velocity of 343 m/s or, more likely, the 10 metre range is a mistake on the part of the person posing the question. 343 m/s is the speed of sound. Fire a bullet at that speed and it would certainly have range longer than 10 m.

Yeah, I know for certain the range is greater than 10 m., so I'll probably mess around with that some more, maybe find the range of a typical RPG and go from there.