physics homework (need help)

if i shoot a gun at exact sea level it leaves the barrel at 450 m/s (meters per second)
at what speed will it hit the ground? (ignore air resistance and wind)



using some formulas in my text i have determined that the bullet will reacha height of 5165 meter's before it stops and begins to fall

ok if it is fire horzontal, which I take it is, you don't say. then the hight would be nothing, and it will hit the the ground at the same speed it is launched from.

vertical

well the speed would be the same as it was fired. Can't think of the equaltion for the height off the top of my head.




reminds me, physic final tuesday.....

what equation did you use?

here


examples 1 & 2 under the THEORY

using the initial velocity (v) 450 m/s we can solve for height (h)

I remember somethat the speed of which a item decelerated upward is the same at which it accelarates, assuming there is no air resistance or wind, henseforth if the item leaves the barrel at 450 m/s and has a deceleration of 9.8 m/s/s it will reach a peak and then fall at 9.8 m/s/s and be travelling at the same velocity as it was shot at the point at which it was shot from. This is the same idea with arrows.

Perhaps I am wrong as it has been like 4 years since I took physics

that seems right but isnt their some maximum speed any feefalling object can reach?

whoa whoa whoa boys. There is no way a bullet can regain the speed at which it was fired by free fall. If that was the case, what do you think would happen every time it hailed? I've been hit by marble sized hail before. If your theory was correct, it would go through my skull and kill me.

In this instance, as stated before, a = g at 9.81m/s^2. The first part of the equation is going to give you the height. I am not going to check your math, so we can assume that you got it. Now you forget about the first part, and solve for the velocity assuming an initial velocity of 0.

First you have to solve for delta t using this equation

Delta S = 1/2a * (Delta t)^2

You already know Delta S, and you know what a is, so you can get t.

Once you're finished there you should already know how to calculate the final velocity. Remember that you're talking in negative terms. (Negative because its falling.)

Hope this helps.

not sure what formula your using..


delta what??

dS = .5(9.8m/s/s) * (dT)^2

Then Vf = a * T (the actual formula is Vf = Vi + (a*t) but the initial velocity is zero since its a free fall equation.

If your height is correct you should be able to get the time using your calculator and that equation. I gotta go to bed, so I hope that addition helps. Otherwise consult your text book.

Only if air resitance and winds affect the speed. You said these were not to be counted.

so in a controlled enviroment the starting velocity equals the end velocity. makes perfect sense when u think of it.



potential energy = mass(m)*gravity(g)*height(h)

kinetic energy = 1/2(mass(m)*velocity^2(v^2))

1/2(v^2)=gh



h= 1 v^2
....2 g

h= 202500
....19.6

h= 10331.6

well i was gonna prove that my maximum height was right and yours wrong, but when i work it out here i see i was in err


if bullets really travel that fast then that means a bullet (ignoring air resistance ) could travel as much as 6 miles high