The equation would still apply as it relies mainly on gravity. and the object by not in contact with any solids so in outer space yes there wont be any atmosphere but the factors that have two be present is the speed of the bullet, the gravitational force of the earth at the point of firing and the angle of firing , the time the ending speed and the displacement all relies on those three factors. In outer space only the gravitational force would change, so the equation would still be applicable.
We will use the following equation of motion to get the answer:
S = (v^2 -u^2)/2*g
where, S = Perpendicular distance traveled from the surface of the earth.
u = muzzle velocity of the bullet
v = final velocity (zero in our case)
g = acceleration due to gravity
So if the muzzle speed of the bullet was 700m/s then,
S = (0 - 700^2)/2*(-9.8) = 25000 meters
So if the bullet was fired at a speed of 700 meters per second than it would go to a height of 25 kilometers.
ASSUMPTIONS:
Thanks @harshcash! But will this equation apply if the bullet gets to outer space??
The equation would still apply as it relies mainly on gravity. and the object by not in contact with any solids so in outer space yes there wont be any atmosphere but the factors that have two be present is the speed of the bullet, the gravitational force of the earth at the point of firing and the angle of firing , the time the ending speed and the displacement all relies on those three factors. In outer space only the gravitational force would change, so the equation would still be applicable.
Thanks @omegagamerow. What happens if the gravitational force on the bullet becomes zero , due to distance from the earth?