General Discussion
In reply to the discussion: RESIDENTS OF OCOTILLO AND BOULEVARD SPEAK OUT, SHARE SAFETY FEARS AFTER TURBINE BLADE FALLS [View all]Chan790
(20,176 posts)I can settle this by telling you the minimum initial-velocity the turbine blade would have had to be moving to travel a horizontal range of 2200m...which is one mile. (I don't have the physics either but I have an online trajectory solver and it has the physics and does the computations for me.) http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra4
v0=R√g/2h
Virtually every other vector in this equation is either known or a constant. There are further factors to consider such as the weight of the blade and the blade has drag as it travels through the air (physics is so much easier in a vacuum)...but all those only act to reduce the range of the blade requiring yet-higher initial velocities...and I suspect that even in ideal circumstance in a vacuum we're nowhere near the needed v0 to propel the blade 2200m.
range (R) is 2200m.
gravitational constant (g) is 9.8m/s2
height (h) is what I just asked you for.
Solving for v0 in m/s
As an example, if the turbine were 200m tall, the blade would have needed an initial velocity along its trajectory of 344.35m/s to travel the needed distance. That's 770MPH.