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

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SUMMARY

The maximum height a .22-caliber bullet, with a muzzle speed of 360 m/s, would reach when fired straight up without air resistance is 6,612 meters. This calculation is derived using the formula h = V(y)^2 / (2g), where V(y) is the initial vertical velocity and g is the acceleration due to gravity (9.80 m/s²). The discussion confirms that the initial setup and calculations are correct, emphasizing the significant height achievable without air resistance.

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  • Understanding of basic physics concepts, specifically kinematics.
  • Familiarity with the equations of motion, particularly h = V(y)^2 / (2g).
  • Knowledge of gravitational acceleration (g = 9.80 m/s²).
  • Ability to perform calculations involving units of meters and seconds.
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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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