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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} ).
d is the total horizontal distance travelled by the projectile. v is the velocity at which the projectile is launched; g is the gravitational acceleration—usually taken to be 9.81 m/s 2 (32 f/s 2) near the Earth's surface; θ is the angle at which the projectile is launched; y 0 is the initial height of the projectile
The vertical component of the velocity on the y-axis is given as = while the horizontal component of the velocity is = . There are various calculations for projectiles at a specific angle : 1. Time to reach maximum height.
Terminal velocity is the maximum speed attainable by an object as it falls through a fluid (air is the most common example). It is reached when the sum of the drag force ( F d ) and the buoyancy is equal to the downward force of gravity ( F G ) acting on the object.
Maximum height can be calculated by absolute value of in standard form of parabola. It is given as H = | c | = u 2 2 g {\displaystyle H=|c|={\frac {u^{2}}{2g}}} Range ( R {\displaystyle R} ) of the projectile can be calculated by the value of latus rectum of the parabola given shooting to the same level.
Escape speed at a distance d from the center of a spherically symmetric primary body (such as a star or a planet) with mass M is given by the formula [2] [3] = = where: G is the universal gravitational constant (G ≈ 6.67 × 10 −11 m 3 ⋅kg −1 ⋅s −2 [4])
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.
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.