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For projectiles in unpowered flight, its velocity is highest at leaving the muzzle and drops off steadily because of air resistance.Projectiles traveling less than the speed of sound (about 340 m/s (1,100 ft/s) in dry air at sea level) are subsonic, while those traveling faster are supersonic and thus can travel a substantial distance and even hit a target before a nearby observer hears the ...
A bullet is a kinetic projectile, a component of firearm ammunition that is shot from a gun barrel. ... (2,730 mph). A bullet from a 9 mm Luger handgun, ...
To calculate the velocity of the bullet given the horizontal swing, the following formula is used: [9] = where: is the velocity of the bullet, in feet per second; is the mass of the pendulum, in grains; is the mass of the bullet, in grains
The same terminal speed is reached for a typical .30-06 bullet dropping downwards—when it is returning to the ground having been fired upwards or dropped from a tower—according to a 1920 U.S. Army Ordnance study. [5] Competition speed skydivers fly in a head-down position and can reach speeds of 150 m/s (490 ft/s).
More simply, the speed of sound is how fast vibrations travel. At 20 °C (68 °F), the speed of sound in air, is about 343 m/s (1,125 ft/s; 1,235 km/h; 767 mph; 667 kn), or 1 km in 2.91 s or one mile in 4.69 s. It depends strongly on temperature as well as the medium through which a sound wave is propagating.
This is a compilation of published detonation velocities for various high explosive compounds. Detonation velocity is the speed with which the detonation shock wave travels through the explosive.
The general formula for the kinetic energy is =, where v is the velocity of the bullet and m is the mass of the bullet. Although both mass and velocity contribute to the muzzle energy, the muzzle energy is proportional to the mass while proportional to the square of the velocity. The velocity of the bullet is a more important determinant of ...
Let m b and v b stand for the mass and velocity of the bullet, the latter just before hitting the target, and let m t and v t stand for the mass and velocity of the target after being hit. Conservation of momentum requires m b v b = m t v t. Solving for the target's velocity gives v t = m b v b / m t = 0.016 kg × 360 m/s / 77 kg = 0.07 m/s = 0 ...