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2.7% (1 in 37) – Picton- Elevation, New Zealand; 2.65% (1 in 37.7) – Lickey Incline, UK; 2.6% (1 in 38) – A slope near Halden on Østfold Line, Norway – Ok for passenger multiple units, but an obstacle for freight trains which must keep their weight down on this international mainline because of the slope. Freight traffic has mainly ...
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
W = walking velocity [km/h] [2] dh = elevation difference, dx = distance, S = slope, θ = angle of slope (inclination). The velocity on the flat terrain is 5 km / h, the maximum speed of 6 km / h is achieved roughly at -2.86°. [5] On flat terrain this formula works out to 5 km/h.
The second equation is identical to the one used to find the weighted average at R / 4; add N × (R/2) where R is the range in feet to the chord average retardation coefficient at midrange and where N is the slope constant factor. [23]
Trajectories of projectiles launched at different elevation angles but the same speed of 10 m/s in a vacuum and uniform downward gravity field of 10 m/s 2. Points are at 0.05 s intervals and length of their tails is linearly proportional to their speed.
The first equation shows that, after one second, an object will have fallen a distance of 1/2 × 9.8 × 1 2 = 4.9 m. After two seconds it will have fallen 1/2 × 9.8 × 2 2 = 19.6 m; and so on. On the other hand, the penultimate equation becomes grossly inaccurate at great distances.
In air, which has a kinematic viscosity around 0.15 cm 2 /s, this means that the product of object speed and diameter must be more than about 0.015 m 2 /s. Unfortunately, the equations of motion can not be easily solved analytically for this case. Therefore, a numerical solution will be examined.
Equation 3 can be substituted in Equation 2. The resulting equation can then be solved for x assuming that y = 0 {\displaystyle y=0} and t ≠ 0 {\displaystyle t\neq 0} , which produces Equation 4.