Angle of projection for maximum range

In summary, the conversation discusses how to find the angle of projection for maximum range when the projectile is shot onto an inclined plane. The formulas (pi/4 - Beta/2) and (pi/4 + Beta/2) are mentioned as well as the use of differentiation to find the maximum range.
  • #1
Abhishek332211
20171006_175217.png
Hi,
I can't seem to work out how angle of projection for maximum range comes out to be (pi/4 - Beta/2)

This happens when the projectile is protected up the inclined plane.

Similarly, I couldn't understand how the angle comes out to be (pi/4 + Beta/2) when the projectile is protected down the inclined plane.

And the other two formulas that have been marked off.

Thanks in advance
 
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  • #2
When the projectile is shot onto a flat ground, do you know how to find the angle for maximum range?

The method is identical here. You are given "R", the range of the projectile. Only this time, you are dealing with two angles, the angle for the initial velocity, and the angle of the inclined plane. For a fix angle of the inclined plane, how would you use "R" to find maximum "R" with respect to the projectile initial angle?

Zz.
 
  • #3
Have you studied differentiation yet?
 

Related to Angle of projection for maximum range

1. What is the angle of projection for maximum range?

The angle of projection for maximum range is 45 degrees. This angle is also known as the optimal angle or the golden angle.

2. Why is 45 degrees the optimal angle for maximum range?

This angle allows for the perfect combination of vertical and horizontal velocity, resulting in the longest possible distance traveled by the projectile.

3. Can the angle of projection for maximum range vary for different objects?

Yes, the angle of projection for maximum range can vary depending on the initial velocity, mass, and shape of the object. However, for most objects, 45 degrees is still the optimal angle.

4. Is there a mathematical formula for calculating the angle of projection for maximum range?

Yes, the formula is: theta = arctan(height/2*range). This formula takes into account the vertical and horizontal components of the initial velocity.

5. Can the angle of projection for maximum range be greater than 45 degrees?

No, the angle of projection for maximum range cannot be greater than 45 degrees. Any angle greater than 45 degrees will result in a shorter distance traveled due to a decrease in the horizontal velocity component.

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