Physics question (not for school)

[Shadow_Storm56]

If I drop a rock and it falls through the air for 3-4 seconds how far will it have gone at 3 and 4 seconds. 

[dalekphalm]

23 minutes ago, Shadow_Storm56 said:

If I drop a rock and it falls through the air for 3-4 seconds how far will it have gone at 3 and 4 seconds. 

You cannot answer the question given the information you've provided.

 

Essentially, you'd need to know the Terminal Velocity of the object.

 

You would need to know the size and weight of the object, it's surface area, and it's drag coefficient, to do all the math.

 

For example, a Skydiver, belly first, has a terminal velocity of about 200 km/h (55.5 m/s). So if the rock was travelling this speed, it would have travelled between 150 and 200 meters.

 

However, a Skydiver can increase their velocity to near 250-290 km/h (69-80 m/s) by nose-diving. This would mean that the rock would travel between 207 and 320 meters in 3 to 4 seconds.

 

Finally, an experienced Skydiver can bring their speed upwards of 480 km/h. I'll stop doing the math, but you get the idea.

 

You can calculate Terminal Velocity as follows:

Quote

where

• Vt{\displaystyle V_{t}}  represents terminal velocity,
• m{\displaystyle m}  is the mass of the falling object,
• g{\displaystyle g}  is the acceleration due to gravity,
• Cd{\displaystyle C_{d}}  is the drag coefficient,
• ρ{\displaystyle \rho }  is the density of the fluid through which the object is falling, and
• A{\displaystyle A}  is the projected area of the object.

 

[Blasteque]

Assuming no resistance:

.5 x 9.8m/ss x 3s x 3s = 44.1m

.5 x 9.8m/ss x 4s x 4s = 78.4m

As for resistance, as @dalekphalm noted above, it gets complicated. 

[Dash Lambda]

2 hours ago, dalekphalm said:

-snip-

I assume he means at the 3 and 4 second marks after dropping it, for which the equation would be this:

x'' = g - (C_dAp/2m)(x')²

Where x is displacement, g is gravitational acceleration, C_d is the drag coefficient, A is the exposed area, m is mass, and p is the fluid density. Substituting in assumptions it'd be this:

x'' = 9.8 - .6(C_dA/m)(x')²

Edited June 12, 2018 by Dash Lambda
Whoops, forgot mass.

[dalekphalm]

35 minutes ago, Dash Lambda said:

I assume he means at the 3 and 4 second marks after dropping it, for which the equation would be this:

x'' = g - (C_dAp/2m)(x')²

Where x is displacement, g is gravitational acceleration, C_d is the drag coefficient, A is the exposed area, m is mass, and p is the fluid density. Substituting in assumptions it'd be this:

x'' = 9.8 - .6(C_dA/m)(x')²

That's true. I simplified it by assuming instant acceleration to terminal velocity, which is technically incorrect.

[Peskanova]

from sea level?  

[Christophe Corazza]

And that’s the difference between a physicist and an engineer ;-)

[wasab]

 

 

On 6/11/2018 at 5:19 PM, Shadow_Storm56 said:

If I drop a rock and it falls through the air for 3-4 seconds how far will it have gone at 3 and 4 seconds. 

This sounds like a typical high school physics problem to me. You sure it ain't your homeowrk?

 

Well, I get these from my college textbooks

a =		dv dt
dv =		a dt
v		t
⌠ ⌡	dv =	⌠ ⌡	a dt
v0		0
v − v0 =		at
v =		v0 + at

Since the derivative of the displacement function is the veclocity function, we can solve its differential equations to get the displacement

 

v =		ds dt
ds =		v dt = (v0 + at) dt
s		t
⌠ ⌡	ds =	⌠ ⌡	(v0 + at) dt
s0		0
s − s0 =		v0t + ½at2
s =		s0 + v0t + ½at2

 

So by the last kinetmatic equation, the distance is just the initial distance plus the initial velocity multiply by time plus 1/2 the acceraleration multiply by time square. 

 

In your case, I assume air resistance is negligible and the fall occurs on planet Earth near the surface where acceraleration due to gravity remains constant.

 

Then distance is just 0.5(9.81m/s)(4*4s)

If you want 3 seconds, just replace the 4.

[Christophe Corazza]

2 minutes ago, wasab said:

This sounds like a typical high school physics problem to me. You sure it ain't your homeowrk?

 

Of course not ;-)

OP mentioned it explicitly in the title, so then it must be true ;-)

[wasab]

14 hours ago, dalekphalm said:

You cannot answer the question given the information you've provided.

 

Essentially, you'd need to know the Terminal Velocity of the object.

 

You would need to know the size and weight of the object, it's surface area, and it's drag coefficient, to do all the math.

 

For example, a Skydiver, belly first, has a terminal velocity of about 200 km/h (55.5 m/s). So if the rock was travelling this speed, it would have travelled between 150 and 200 meters.

 

However, a Skydiver can increase their velocity to near 250-290 km/h (69-80 m/s) by nose-diving. This would mean that the rock would travel between 207 and 320 meters in 3 to 4 seconds.

 

Finally, an experienced Skydiver can bring their speed upwards of 480 km/h. I'll stop doing the math, but you get the idea.

 

You can calculate Terminal Velocity as follows:

 

You are over thinking it....

3-4 sceonds ain't enough to hit terminal velocity, not for a rock on Earth anyways.

[Dash Lambda]

15 hours ago, Erik Sieghart said:

You guys go drop a large rock and a small rock off a bridge (so it falls for 3 seconds) and tell me how much air resistance affected it (here's a hint, negligibly).

 

Just answer the question as though there wasn't any and you'll have a good enough estimation.

He did say "through the air," I always take that sort of thing to mean it's part of the problem.

 

OP, the solution without air resistance is a lot simpler:

Your acceleration, the second derivative of position, is g, so x''=g.

Integrate twice: x' = gt, x = (g/2)t²

 

Constants of integration are assumed to be 0 because you're looking for the change in position from a standstill.

[Canada EH]

On 6/11/2018 at 3:19 PM, Shadow_Storm56 said:

If I drop a rock and it falls through the air for 3-4 seconds how far will it have gone at 3 and 4 seconds. 

 

On 6/11/2018 at 4:26 PM, Blasteque said:

Assuming no resistance:

.5 x 9.8m/ss x 3s x 3s = 44.1m

.5 x 9.8m/ss x 4s x 4s = 78.4m

As for resistance, as @dalekphalm noted above, it gets complicated. 

There is always resistance!

[Canada EH]

6 minutes ago, Erik Sieghart said:

Are you also going to calculate all of your other values to 10 digits so your air resistance will actually make the numbers different?

 

Better make sure to also include the gravitational force of the moon.

What about altitude?

[Christophe Corazza]

5 hours ago, Canada EH said:

What about altitude?

 

5 hours ago, Erik Sieghart said:

yeah don't forget that

 

20 hours ago, wasab said:

You are over thinking it....

3-4 sceonds ain't enough to hit terminal velocity, not for a rock on Earth anyways.

 

Guys... Can't we simply solve this by Monte Carlo simulation? 

[grimreeper132]

Assuming

1. No air resistance
2. It is near the surface of the earth and the slight fluctuations in gravity can be ignored so gravity can be assumed to be 9.81 m/s²
3. The rock doesn't reach terminal velocity or relativistic speeds
4. The rock is dropped perfectly, with no horizontal, vertical or rotational velocities (starts at complete rest)
5. The rock is a uniform body with it's weight distribution equal throughout it and so there is no moments/torques acting on the rock causing it to rotate
6. There is no energy loss to the surroundings and the rock perfectly with 0 losses transfers it's potential to kinetic energy
7. The rock is high enough that it does not impact anything as it falls
8. The rotation of the body it is falling towards (in this case it is assumed to be earth) can be ignored
9. There is NO other external forces that are acting on the rock and any that are can be ignored
10. The equations V=u+at and S=ut+1/2at² hold true for this example and can be used to estimate a velocity

Velocities after this time

Then for 3s

V=u+at

V=0+9.81*3

V=29.43 m/s

 

For 4s

V=u+at

V=0+9.81*4

V=39.24 m/s

 

Distance at 3s

S=ut+1/2at²

S=0*3+ 1/2*9.81*3²

S=44.1m

 

Distance at 3s

S=ut+1/2at²

S=0*4+ 1/2*9.81*4²

S=78.5m

 

If you do not want to assume air resistance is negligible then have fun as I can't be bothered with dealing with a force which is changing with time, as that requires effort to work out and do, and fuck that, I already do enough of that at work

[grimreeper132]

4 hours ago, Erik Sieghart said:

The rocks center of mass doesn't matter.

 

it does because by having a rock that's centre of gravity not being in the centre of the body this can cause a moment, thus causing it to rotate slightly, thus losing potential energy to rotation

[wasab]

3 hours ago, grimreeper132 said:

it does because by having a rock that's centre of gravity not being in the centre of the body this can cause a moment, thus causing it to rotate slightly, thus losing potential energy to rotation

Huh? What? I thought moment of inertia means pivoting about an axis like in the case of seasaw where like say gravity on one end will apply a torque causing it to rotate about the center of mass, the pivoting point in this case. I don't see how a falling rock will rotate on its own for just falling on planet earth.

 

Either my brain is shot or what I learned in ap physics mechanics is dead wrong. 

 

Edit: I see what you are talking about. It is true if an object is held at its center of mass, it won't rotate due to gravity because weight is distributed uniformly however what you are suggesting here is that earth will start to rotate because ball isn't at the center of mass which in this case is the center of earth because of the mass of earth compare to the rock. This is just ridiculous. 

[WolfLoverPro]

On 11/06/2018 at 10:19 PM, Shadow_Storm56 said:

If I drop a rock and it falls through the air for 3-4 seconds how far will it have gone at 3 and 4 seconds. 

It will go as far as it can until it hits the earth

[dalekphalm]

1 hour ago, Erik Sieghart said:

No because gravity isn't acting on an axis; it's acting on all parts of the rock equally.

Technically this isn't correct either, since gravity will act on the nearest part of the rock the strongest, and the farthest part of the rock the weakest.

 

Mind you, the difference in the case of a rock is probably undetectable or near zero.

[wasab]

2 hours ago, dalekphalm said:

Technically this isn't correct either, since gravity will act on the nearest part of the rock the strongest, and the farthest part of the rock the weakest.

 

Mind you, the difference in the case of a rock is probably undetectable or near zero.

Not if it is falling into a black hall. The difference will be great enough to stretch its atoms into a long line of spaghetti

[Christophe Corazza]

6 hours ago, wasab said:

Not if it is falling into a black hall. The difference will be great enough to stretch its atoms into a long line of spaghetti

 

I think OP is still talking about dropping a rock on Earth though 

[Dash Lambda]

On 6/13/2018 at 10:35 AM, Erik Sieghart said:

Horizontal forces do not affect downward acceleration. They're independent vectors.

Actually, since the Earth is a sphere, horizontal forces would push it away from the planet. That's how satellites stay in orbit -They're constantly in freefall, but they're moving parallel to the surface fast enough that it cancels out with the inward acceleration.

And as for rotation, tumble makes the aerodynamics even uglier.

 

... Please forgive me.

 

On 6/13/2018 at 9:03 PM, dalekphalm said:

Technically this isn't correct either, since gravity will act on the nearest part of the rock the strongest, and the farthest part of the rock the weakest.

 

Mind you, the difference in the case of a rock is probably undetectable or near zero.

Well, let's play around with Newtonian gravity...

 

Radius of the Earth: ~6,371,000 m

Now let's say a hypothetical basket-ball sized (.2426m diameter) rock is dropped 2m above the surface, the ratio between the forces reduces to

F_L/F_H = (6371002.2426 / 6371002)^2 =~ 1.000000076

 

So, in Newtonian gravity, the lower edge is pulled on the order of 0.00001% harder than the top edge. Actually more than I thought. For a 1km tall building, it's ~0.03%, which is a hell of a lot more than I thought.

[Unhelpful]

At least one foot.

[wasab]

15 hours ago, Dash Lambda said:

Actually, since the Earth is a sphere, horizontal forces would push it away from the planet. That's how satellites stay in orbit -They're constantly in freefall, but they're moving parallel to the surface fast enough that it cancels out with the inward acceleration.

And as for rotation, tumble makes the aerodynamics even uglier.

 

... Please forgive me.

 

Well, let's play around with Newtonian gravity...

 

Radius of the Earth: ~6,371,000 m

Now let's say a hypothetical basket-ball sized (.2426m diameter) rock is dropped 2m above the surface, the ratio between the forces reduces to

F_L/F_H = (6371002.2426 / 6371002)^2 =~ 1.000000076

 

So, in Newtonian gravity, the lower edge is pulled on the order of 0.00001% harder than the top edge. Actually more than I thought. For a 1km tall building, it's ~0.03%, which is a hell of a lot more than I thought.

If it has horizontal velocity, its traiectroy will be a parabola. If its horizontal velocity is so great, it will outright circle the planet Earth and hence never falls. The gravity will become a centripetal force in such case. 

[TheRandomness]

Using the simplest (and only) equations I know (because A-level :P) you can just assume air resistance is negligible, initial velocity is 0ms^-1 and acceleration is 9.81ms^-2, therefore you can use suvat stuff to get the distance.

s = ut + (at²)/2, where s is displacement, u is initial velocity, t is time and a is acceleration

s = (9.81*3²)/2 = 44.145 for up to 3 seconds after drop

s = (9.81*4²)/2 = 78.48 for up to 4 seconds after drop
So it travels 34.335m in the 3rd second of free fall assuming negligible air resistance and a vertical drop.

 

I think this is the first time I've done anything related to physics equations outside of school