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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.
American Forests, for example, uses a formula to calculate Big Tree Points as part of their Big Tree Program [3] that awards a tree 1 point for each foot of height, 1 point for each inch (2.54 cm) of girth, and ¼ point for each foot of crown spread. The tree whose point total is the highest for that species is crowned as the champion in their ...
θ is the angle at which the projectile is launched; y 0 is the initial height of the projectile; If y 0 is taken to be zero, meaning that the object is being launched on flat ground, the range of the projectile will simplify to: =
where is the superelevation in inches, is the curvature of the track in degrees per 100 feet, and the maximum speed in miles per hour. The maximum value of cant (the height of the outer rail above the inner rail) for a standard gauge railway is approximately 150 mm (6 in).
Then calculate the central angle in radians between two points (,) and (,) on a sphere using the Great-circle distance method (haversine formula), with longitudes and being the same on the sphere as on the spheroid.
A set of equations describing the trajectories of objects subject to a constant gravitational force under normal Earth-bound conditions.Assuming constant acceleration g due to Earth's gravity, Newton's law of universal gravitation simplifies to F = mg, where F is the force exerted on a mass m by the Earth's gravitational field of strength g.
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This formula allows one to find the angle of launch needed without the restriction of =. One can also ask what launch angle allows the lowest possible launch velocity. This occurs when the two solutions above are equal, implying that the quantity under the square root sign is zero.