How high ?
#1
Thread Starter
Fork Horn
Joined: Jan 2004
Posts: 269
Likes: 0
From: Finland
Here is dilemma for all of you who don't have nothing else to do? or who just can't refuce matematic task. which is nearly same thing.
You shoot 8g .308 bullet 800 m/s straight up. How high it goes? (don't mind about weather or sunlight
.
You shoot 8g .308 bullet 800 m/s straight up. How high it goes? (don't mind about weather or sunlight
.
#3
Well, if you don't take the effect of drag into account, and assume the force of gravity to be constant at 9.8m/s^2, then a bullet fired at 800m/s straight up would reach a max altitude of roughly 16,327m in 40.8 seconds. Total flight time is roughly 82 seconds.
If you have a graphing calculator (Texas Instruments TI-83 or similar) you can use the following parametric function to calculate the trajectory of a projectile sans drag(make sure the calculator is in degree and parametric graph modes). I do not (yet) know the function(s) necessary for accurately calculating the effect of drag on the bullet. I haven't gotten that far in calculus yet.
X(t)=(velocity in m/s)t*cos (launch angle in degrees)
Y(t)=(velocity in m/s)t*sin (launch angle in degrees) - 9.8t^2
So the functions I used for your calculation are:
X(t)=0 (because there is no horizontal component of velocity)
Y(t)=800t*sin 90 - 9.8t^2
Mike
If you have a graphing calculator (Texas Instruments TI-83 or similar) you can use the following parametric function to calculate the trajectory of a projectile sans drag(make sure the calculator is in degree and parametric graph modes). I do not (yet) know the function(s) necessary for accurately calculating the effect of drag on the bullet. I haven't gotten that far in calculus yet.

X(t)=(velocity in m/s)t*cos (launch angle in degrees)
Y(t)=(velocity in m/s)t*sin (launch angle in degrees) - 9.8t^2
So the functions I used for your calculation are:
X(t)=0 (because there is no horizontal component of velocity)
Y(t)=800t*sin 90 - 9.8t^2
Mike
#5
Nontypical Buck
Joined: Feb 2003
Posts: 3,393
Likes: 0
From: Western Nebraska
I agree with driftrider on the 82 seconds, but the average speed of 400m/sec times 82 is about double the height he has predicted.....I say the bullet reaches 32,800 meters.
The weight of the bullet isn't relevant for this type of calculation where air resistance is neglected.
The weight of the bullet isn't relevant for this type of calculation where air resistance is neglected.
#9
Guest
Posts: n/a
Hey smartypants. Now add .277 Drag Coef with some air. Neqlecting that air is lighter the higher you go and neglecting the fact that BC changes over velocity. For a 8gram .308 bullets sierra calls for .277 BC less than 2800 but more than 2000fps. Let see who gets it first.
#10
Typical Buck
Joined: Jan 2004
Posts: 549
Likes: 0
From:
if a 30-06 will travel 5miles then the 308 will go atleast 4.5 miles. But I'm sure you guys will let us know the exact yardage,feet, inches, and how long it took to hit the ground again. lol good post I got a laugh on it..



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