I came up with 6828.16 meters. For total height. Now I haven't used my physics in 6 years but I used to be pretty proficient at it.

Calc time for height we know that the starting velocity 800m/s and the final velocity is 0 m/s(just before it starts coming down). Acceleration is a constant -9.8m/s^2. We get 0=800m/s-9.8m/s^2*t^2 which is (800m/s)/(9.8m/s^2)=t^2 which means t=9.035s.

Now distance is D=InitialVelocity*t+.5*acceleration*time^2. D=(800m/s)*(9.035s)+.5(-9.8m/ss)*(81.6ss) D=6828.16


Calculating drag depends on air density. Since that changes with altitude that is just way to much to go into.

Sloth,

You're answer is incorrect because in your first step you are confusing the t^2 term in the position function with the t term in the velocity function. In your first step you incorrectly used t^2 rather than t, which causes your units for the answer to come out wrong. In your version after solving for t and paying close attention to units the answer comes out to t=9.035 (seconds^(1/2)) Or, in simpler terms the units are the square root of seconds. To get the desired units we'd have to square the answer only to get t=81.6 seconds.

The vertical position as a function of time is a parabola, implying a quadratic function which is given as h(t)=-4.9t^2-800t.

The reason for this is given in my explanation above and involves working backward to find the position function by knowing that the velocity function v(t) is the derivative of the position function h(t), and that the acceleration function a(t) is the second derivative of h(t), or the derivative of v(t). Working backward as I explained above gives us the following functions that describe the motion of the bullet (sans drag) as a function of time t.

Acceleration a(t)=-9.8m/s^2 (this is a constant function because accelleration is contant regardless of time)

Velocity v(t)=A(t) (the antiderivative of a(t))=h'(t)= -9.8 (m/s^2) X t (sec.) + 800(m/s) if you work this out you'll see that the units cancel out to give the answer in meters/sec., which is a unit of velocity that we're looking to find with this function.

Height (in meters m) h(t)=V(t)=-4.9(m/s^2) x t(s)^2 + 800(m/s) x t(s) + 0(m) (the last term is the starting position, which we're calling zero since we consider the muzzle the zero)

Solving the velocity function for zero gives us t=81.6 seconds when v(t)=0.

Plugging t=81.6seconds into the height function h(t) we find that the height in meters when velocity is zero is 32,653 meters [h(81.6)=32,653m].

Mike

Sloth, you are right about the difficulties of including drag. For one thing, you would have to know air density (temperature and altitude, AKA density altitude) at the time the shot is fired, and also the ballistic coefficient of the particular projectile used. However, it seems to me that drag is THE major factor that causes a bullet to lose forward momentum unless it hits something more substantial than air. Therefore, to ignore it is to overlook the major limiting factor on the vertical distance any gun can shoot. Early ballisicians had a devil of a time trying to produce firing tables for cannon because they were not certain the extent to which air resistance affected their projectiles, until they invented the ballistic pendulum and were able to determine cannon ball velocities near the muzzle and at extended distances. They were just amazed to discover just how much velocity was lost due to drag.

Good job drifter. I really blew that one. I take it your still in college, so when you get out don't forget all that stuff you learned. I actually have a physic degree and a math minor, but 6 years of not using either one has really trashed my skills.

All I can say is, dont be shooting in the air. You never know what it will hit on the way down.

I teach physics, and my calculations agree with bigcountry. That is excluding the friction coeficent of the different air density as the bullet reaches higher altitudes.