2007-11-17 09:42:28 · 4 answers · asked by whoopsidaisy 3 in Science & Mathematics ➔ Physics
Neglecting air friction, the equation of motion of the bullet is
y(t) = vo*t - (1/2) g t^2 = t (vo - g*t/2). (vo is the initial speed)
Then, it will last 2 vo / g in the air.
I have no idea about bullet speed, but just for obtaining a number I will put vo=1km/s, then t = 2x10^3/10 sec = 200 sec, more than 3 minutes.
2007-11-17 09:57:06 · answer #1 · answered by GusBsAs 6 · 0⤊ 0⤋
Velocity for different guns and bullets vary widely, so there is no one answer.
There is a certain speed that friction of the air restricts any further acceleration. This is called terminal velocity. this will vary by the size and shape of the falling object.
2007-11-17 09:55:18 · answer #2 · answered by Jeffery H K 6 · 0⤊ 0⤋
t = 2 * v / g
t = the time (s)
v = initial velocity (m/s) (speed of the bullet as it leaves the barrel)
g = gravitational acceleration (9.8 m/s/s)
* air friction was ignored
2007-11-17 09:54:47 · answer #3 · answered by nf119 3 · 0⤊ 0⤋
it depends on what kind of gun and bullet.
2007-11-17 09:49:50 · answer #4 · answered by Shannan M 2 · 0⤊ 0⤋