Maximizing Projectile Distance: Calculation at 45 Degrees

AI Thread Summary
The maximum distance a projectile can travel occurs at a launch angle of 45 degrees due to the relationship expressed in the range formula, R, which includes the term sin(2α). This indicates that the range is maximized when 2α equals 90 degrees, leading to α being 45 degrees. The discussion also touches on the possibility of deriving this result from kinematic equations, suggesting a deeper exploration of the physics involved. The participants express a willingness to share calculations and insights on the derivation process. Overall, the key takeaway is the mathematical basis for maximizing projectile distance at a 45-degree angle.
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When you kick a ball of mass m into the air at a speed v0 at an angle theta to the ground, how do you know that the maximum distance in the x direction that the ball can travel occurs at 45 degrees.

Can someone show me the calculation of how this happens?
 
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The formula for the range, R, of a projectile has a term

\sin(2\alpha)

where \alpha is the launch angle above the horizon. This means that the range will be a maximum when the projectile is launched such that

2\alpha = 90^o
 
thanks! how did u get that equation. is it possible to derive it from the kinematics equations? THanks!

nvm i think i did it
 
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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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