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Topic review (newest first)
- 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)}.