How High Would a Bullet Rise if Fired Straight Up Without Air Resistance?

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When a .22-caliber bullet is fired straight up at a muzzle speed of 360 m/s, it would theoretically rise to a maximum height of 6,612 meters without air resistance. The calculation involves using the equation y = (v^2 - initial v^2) / (2a), where 'v' is the final velocity (0 m/s at the peak) and 'a' is the acceleration due to gravity (-9.8 m/s^2). Some participants humorously noted the impracticality of such a height, suggesting a more realistic maximum height would be around 6 kilometers. The discussion also highlighted the importance of calculating the time to reach this height, which can be derived from the initial speed divided by gravitational acceleration. Overall, the calculations confirm that, theoretically, the bullet could reach an extreme altitude if air resistance were not a factor.
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I can't figure out if I set this problem up correctly. Any help?

The muzzle speed of a 22-caliber bullet fired from a rifle is 360 m/s. If there were no air resistance, how high would this bullet rise when fired straight up?

v^2=intial v^2+2ay
y=v^2-initial v^2/2a=(0-(360 m/s)^2)/2(-9.80 m/s^2)=6,612 m


The maximum height reached is 6,612 m.
 
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No, could you imagine a bullet being shot in the air and reaching a maximum heaight of 6.6 meters? lol:P

First find the time it takes...360/9.8...then it should be relatively simple to figure out
 
No, 6km is correct. Youll find a general term for what you described above as:

h = y(peak) = \frac{V(y)^2}{2g}

Where V(y) is the initial vertical component of the velocity
 

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