Finding the Ideal Angle for Maximum Distance with Friction

AI Thread Summary
The discussion focuses on determining the ideal angle for maximum horizontal distance when throwing an object, specifically considering the effects of friction. The established formula for distance without friction is s = v²sin(2θ)/g, which indicates that the optimal angle is 45 degrees. However, the user seeks to derive a new formula that accounts for friction, hypothesizing that the ideal angle will be slightly less than 45 degrees. There is also mention of the observation that aiming below 45 degrees can yield a greater range, regardless of friction. The conversation emphasizes the need for a mathematical proof to support these claims.
vagelier
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Homework Statement


So for physics, I am trying to establish what the ideal angle is to get a maximum horizontal distance when you throw an object. So far, by using the formula s = v²sin(2θ)/g I've discovered that the angle is 45 degress without friction (duh). But now I need to have a formula to calculate the ideal angle with friction. I know it should be a little below 45 degrees, but I still need to prove it by equation.


Homework Equations


s = v²sin(2θ)/g


The Attempt at a Solution


see 1.
 
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vagelier said:

Homework Statement


So for physics, I am trying to establish what the ideal angle is to get a maximum horizontal distance when you throw an object. So far, by using the formula s = v²sin(2θ)/g I've discovered that the angle is 45 degress without friction (duh). But now I need to have a formula to calculate the ideal angle with friction. I know it should be a little below 45 degrees, but I still need to prove it by equation.


Homework Equations


s = v²sin(2θ)/g


The Attempt at a Solution


see 1.

Why would SIN(28) be a constant?

Also, aiming a cannon lower than 45 degrees produces a greater firing range regardless of frictiion.

Have you left things out?
 
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