Does The Range Equation Work for Complementary Angles?

I am a physics teacher and am wondering something about the range equation. Wouldn't that equation work for complementary angles each time? The author of our book has us solving for one angle but shouldn't you solve for two?
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 Recognitions: Science Advisor If given range, you are solving for angle, yes. You should get two solutions or none. The only exception is maximum range, which has precisely one solution, which is 45°. In general, if θ is a solution to a range equation at given range, then so is 90°-θ.

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 Quote by K^2 In general, if θ is a solution to a range equation at given range, then so is 90°-θ.
Unless, of course, it is an inclined plane. If the plane is at angle α above horizontal in the downrange direction, the solutions will be θ and π/2-θ+α.

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Does The Range Equation Work for Complementary Angles?

I never realized that range with elevation change formula can have such an elegant form when change of elevation is expressed as an incline angle. Good to know.

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 Quote by slr20042003 Wouldn't that equation work for complementary angles each time? The author of our book has us solving for one angle but shouldn't you solve for two?
In real setups, the steeper solution is often impractical, as it gives more drag (and might be more difficult to achieve).
 Recognitions: Science Advisor Depends on what you are doing with it. There is a demo I used to show to undergrads in the physics lab. Ask one student to set up a plastic cup 4-6 meters from a spring-loaded cannon (max range ~7m), measure the distance to the cannon, compute angle, shoot a ball bearing into the cup. In that case, you want to go with steeper angle, because it gives you a bigger cross-section to hit and is less likely to knock the cup over. Really drives the point across that simple physics that they learn can actually be used to predict something practical with such precision. Point is, there are two solutions. Whether one is more practical than the other is a separate issue that has nothing to do with problem itself.

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