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The path of this projectile launched from a height y 0 has a range d. In physics, a projectile launched with specific initial conditions will have a range. It may be more predictable assuming a flat Earth with a uniform gravity field, and no air resistance. The horizontal ranges of a projectile are equal for two complementary angles of ...
The range and the maximum height of the projectile do not depend upon its mass. Hence range and maximum height are equal for all bodies that are thrown with the same velocity and direction. The horizontal range d of the projectile is the horizontal distance it has traveled when it returns to its initial height ( y = 0 {\textstyle y=0} ).
They fired projectiles weighing from 1,900 to 2,700 lb (860 to 1,220 kg) at different muzzle velocities, depending on the projectile. When firing armor-piercing projectiles, their muzzle velocity was 2,500 feet per second (760 m/s) with a range of up to 24 mi (39 km). At maximum range the projectile spent almost 1 + 1 ⁄ 2 minutes in flight ...
Range (): The Range of a projectile is the horizontal distance covered (on the x-axis) by the projectile. Mathematically, = ... The Range is maximum when angle ...
At 1,500 m (1,640 yd) range the projectile velocity predictions have their maximum deviation of 10 m/s (32.8 ft/s). The predicted total drop difference at 1,500 m (1,640 yd) is 0.4 cm (0.16 in) at 50° latitude.
A railgun projectile without the ability to change course can hit fast-moving missiles at a maximum range of 30 nmi (35 mi; 56 km). [58] As is the case with the Phalanx CIWS, unguided railgun rounds will require multiple/many shots to bring down maneuvering supersonic anti-ship missiles, with the odds of hitting the missile improving ...
With the addition of clinometers fixed machine gun squads could set long ranges and deliver plunging fire or indirect fire at more than 2,500 m (2,730 yd). This indirect firing method exploits the maximal practical range, that is defined by the maximum range of a small-arms projectile while still maintaining the minimum kinetic energy required to put unprotected personnel out of action, which ...
To find the angle giving the maximum height for a given speed calculate the derivative of the maximum height = / with respect to , that is = / which is zero when = / =. So the maximum height H m a x = v 2 2 g {\displaystyle H_{\mathrm {max} }={v^{2} \over 2g}} is obtained when the projectile is fired straight up.