I Have A Bet On.....

[YT2095]

But because you need the spin to stabilize it' date=' I'd expect the optimal spin to be such that it stopped spinning after it's designed range.

 

Anyway, you originally stated that the upward momentum and spin ceased simultaneously at the zenith of the vertical shot, and I wonder if you meant to imply that, since your answer here doesn't support it.[/quote']

 

that was better defined in my post #21

[Dave]

yes, but they are trying to give a reason why[/i'] it won't work. They just don't give the answer!

 

Try phrasing it this way. Since the guy obviously believes in gravity, liken the situation you're talking about to a man jumping out of a plane, with a parachute. Point out the simple fact that when the guy pulls the rip cord, he doesn't continue accelerating and die in a mess on the floor

 

Either that or just beat it into him

[blike]

Only certain strands are harmful, most of them are harmless. We all have E. Coli living in our intestines (yummy eh).

[blike]

ROFL, wrong thread.

[Dave]

Yup. But yeah, the parachute thing might persuade him to believe that a similar effect is enforced in any object travelling through the air.

[Dudde]

what if E. coli were living in the parachutes O_O

 

and where did parachutes come into this?

[aommaster]

Try phrasing it this way. Since the guy obviously believes in gravity' date=' liken the situation you're talking about to a man jumping out of a plane, with a parachute. Point out the simple fact that when the guy pulls the rip cord, he doesn't continue accelerating and die in a mess on the floor

 

Either that or just beat it into him [/quote']

 

i think u guys got me wrong. I was explaining it to dudde, who said

 

there isn't any reason (nor reasoning) that the bullet would move the same speed, although the rest of the thread is proving this, so you should be fine

[xXxAuroraxXx]

The bullet would come down at less of a speed it did when it was shot up. Because the force of the gun shooting it is gone. As "Sayanora" stated, the bullet stops completly before it begins it's downward fall.

[Dave]

The bullet would come down at less of a speed it did when it was shot up. Because the force of the gun shooting it is gone. As "Sayanora" stated, the bullet stops completly before it begins it's downward fall.

 

If it wasn't for things like air resistance, then the bullet would come down at the same speed that it was fired. Simple kinematics and parabolic paths can tell you this.

[Sayonara]

Is that assuming it does not leave the barrel of the gun along a planetary radius?

[Dave]

I'm talking about this in simple terms, a perfectly flat surface (which the Earth is not), a constant gravitational force (which the Earth doesn't have) and no air resistance.

[xXxAuroraxXx]

If it wasn't for things like air resistance, then the bullet would come down at the same speed that it was fired. Simple kinematics and parabolic paths can tell you this.

 

Well does the strength of air resistance/wind/air pressure surpass the strength of the gun shooting the bullet? If anything it would come down the same speed, but most likely slower.

 

Ehh, and I haven't studied Kinematics and I JUST started sudying parabolas. I'm a freshmen in highschool, give me a break

[Dave]

I'm too tired to think atm; it's quite possible that what I posted was utter nonsense, I'll look it over tomorrow.

[blike]

It's not david (assuming certain conditions).

 

Think about it: The bullet will stop ascending when the force that gravity exerts has finally overcome the force of the charge in the gun. At that point, the bullet's y-velocity is 0 and the bullet is at it's highest altitude. The force of gravity will continue exerting the same force over the same period of time until the bullet hits, and thus, the final velocity will be the negative of the original velocity.

 

intial velocity: y (m/s)

veolicty at apex: 0 m/s

final velocity: -y (m/s)

 

According to Sir Netwon (in an ideal vacuum):

v(t) = v_o + at

 

Let's assume a bullet fired from a gun has an initial velocity of 896m/s and we want to know how much time will pass before the bullet starts to fall again (note that the bullet's velocity = 0 at the peak)

 

0 m/s = 896m/s + -9.8m/s^2(t)

t = 91.43s

 

Thus, the bullet will continue upwards for 91.43 seconds before it starts to fall again. With the same equation, we can set the intial velocity = 0 (starting at the peak), and 91.43s fall time. (because a plot of displacement vs. time is symmetrical for a projectile in a vacuum)

 

v = 0m/s + -9.8m/s(91.43s)

v_final = 896m/s

[alt_f13]

On the moon it would not have a terminal velocity. Now I forgot how to work out escape velocity, but I'm pretty sure your moon bullet would return at the same speed it left.

 

[edit]

 

Tarnation. I missed this last page.. consider this post a recap.

[Radical Edward]

On the moon it would not have a terminal velocity. Now I forgot how to work out escape velocity' date=' but I'm pretty sure your moon bullet would return at the same speed it left.

[/quote']

 

calculate the potential energy the bullet would have at an infinite distance, and then caculate the velocity required to give the bullet an equal kinetic energy.

 

other than that, yes, the bullet would return at he same speed as it left.

[alt_f13]

Oh, right. Thanks.

[swansont]

calculate the potential energy the bullet would have at an infinite distance' date=' and then caculate the velocity required to give the bullet an equal kinetic energy.

[/quote']

 

Not quite. The PE at infinity is zero, using PE=-GMm/r. It's the mechanical energy that is zero at infinity, where ME=KE+PE

 

So you want to find the PE at the surface, and add an equal magnitude of KE.