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scientia
2012-12-27 14:39:50

I found some more errors and corrected them. And now I'm sure everything up there is perfect.

The challenge now is to compute the actual integral. Not for the faint-hearted.

scientia
2012-12-27 13:20:34

I made some careless mistakes in my calculations and got some horrendously frightening values for the elevating angle.

I have re-computed them. They should be

and

And you want to find the following integral for the solid angle:

scientia
2012-12-26 09:30:27

I think you mean
and also you want
.

Supposed the trajectory is angled
horizontally from the vertical plane through the y-axis, taking
to be positive to the right and negative to the left of the y-axis. The straight line on the ground through the origin making this angle with the y-axis intersects the circle at two points. The distances from the origin to these two points can be calculated; these are the minimum and maximum ranges respectively at the angle
. Let
be the vertical angle from the ground at which the particle must be projected to cover the minimum range, and
be the vertical angle to reach the maximum range. (This can be computed from the formula
where
is the acceleration due to gravity and
the desired range.) The maximum value of the horizontal angle
is
.

Then the solid angle you want is:

rajinikanth0602
2012-12-26 00:31:18

Consider X-Y plane as ground and gravitational field acts perpendicular to it.A particle is projectedwith a velocity ''v'' at an angle less than 45°from origin onto circle (x)²+(y-d)²=r² on X-Y plane.Find the solid angle made by all projectionsof particle onto every point on the circle.{range ''R'' of particle≥(d+r)}.