That depends upon whether it is fired into a wall just two feet away (in which case it travels a little over two feet!) or (as the intent of this question assumes NO obstructions) how high above the earth is the gun when discharged? Thus, ignoring friction with the air molecules, the horizontal distance the bullet travels is determined by the VERTICAL COMPONENT of the bullet's trajectory. The vertical component is just the pull of gravity on that bullet!! Here's an example: regardless of the bullet's muzzle velocity or mass first calculate the time it takes for the bullet to 'fall' the distance between the gun's muzzle and the earth's surface at a rate of 4.90 meters per second per second. This is a little confusing but goes like this... it falls 4.90 meters the 1st second; 4.90 x 4 s(i.e 2 seconds squared) after 2 seconds of fall; 4.90 x 9 in 3 seconds of fall = 44.1 meters; etc. So- for simplicity say firing the round 4.90 meters above the earth's surface (assumed horizontal (not curved))it would take the bullet (40 caliber or not) 1 second to reach the ground. Now figure the HORIZONTAL component of that bullet over that same one second time interval... Let's say the bullet travels at 425 meters per second then it would travel 425 meters before reaching the dirt!
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Well, first...there is no such thing as a 40MM pistol. Since 1 inch is about 25MM's, that would make for a pretty large bullet coming out of your gun.
The 40Smith & Wesson cartridge will send a bullet out of the muzzle at around 1000 to 1200 FPS...that is feet per second.