Relationship Between Starting Angle of a Projectile and it's distance

Question

If we say that we shoot a projectile at 100% force, and at 45 degrees and it goes n meters, then if we shoot the same projectile at 50% force and 45 degrees it will go n/2 meters. Now if we halve the angle instead of the force, now shooting the projectile at 100% force and 22.5 degrees, it won't go n/2 meters. I have looked at a few websites detailing the relationship between the initial velocity, and the angle of a projectile leading to its range, but I am still confused about how the starting angle affects the percentage of how far it could've gone.

Can somebody please write an equation to calculate what percent of n a projectile will have gone with varying angles?

Answer 1

Based on your reply to Vasya's answer, is this closer to what you're looking for?

$\dfrac{d(22.5)}{d(45)}=\dfrac{\frac{1}{9.81}v^2 \sin(45^\circ)}{\frac{1}{9.81}v^2\sin(90^\circ)}=\dfrac{\sin(45^\circ)}{\sin(90^\circ)}=\dfrac{\frac{1}{\sqrt{2}}}{1}=\dfrac{1}{\sqrt{2}}\approx 0.707$

Answer 2

There are plenty resources on the internet that describe projectile motion. In case the projectile is launched from the surface with velocity $v$, the distance can be found using this formula: $d=v^2 sin(2θ)/9.81$. Thus, the distance is the greatest when angle is 45 degrees. Reference: http://homepage.usask.ca/~dln136/projectile/pages/module5.html