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2 Meinungen für “How can I find an additional angle that will give the same range as my experimental angle?”

1. kirchwey sagt:
   11.September, 2009 um 3:01 pm

   For a constant launch velocity and a trajectory whose launch and impact elevations are the same, the maximum range occurs at 45 deg elevation. The range decreases continuously from maximum to zero as the elevation goes from 45 to 0 or from 45 to 90. Thus if your first angle is < 45, there must be another angle > 45 that results in the same range. In fact the two angles are complementary, thus your 2nd angle = 70 deg.
   The formula needed, R = v0^2*sin(2theta)/g is derived in the ref. For maximum range R, sin(2theta) must be = 1, so theta = 45 deg.
   References :
   http://hyperphysics.phy-astr.gsu.edu/Hbase/traj.html#tra4

2. knr sagt:
   11.September, 2009 um 3:06 pm

   For a projectile the horizontal range is given by
   R = u^2 sin ( 2xangle of projection )/g
   but from trigonometry, we know that sin (any angle) = sin( 180 – that angle)
   hence, R = u^2 sin (2 angle ) /g = u^2 sin (180 -angle)/g
   = u^2 sin 2 (angle)/g = u^2 sin 2 ( 90 – angle )/g
   thus, we see that for same velocity, range will be same
   for some angle = 90 – the same angle
   e.g., if the angle is 30 degree, the other angle will be 90 – 30 = 60 degree
   , in fact these two angles will be called complementary angles for which the range will be same.
   References :

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