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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?

amWhy
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sebtheiler
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2 Answers2

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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

Vasili
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  • I understand these pages and what they are saying, but I simply want to know the relationship between the range when launched at 45 degrees and the range when launched at for say 22.5 or 10 degrees. I.E., an equation to describe that launching a projectile at 22.5 degrees is about 70 percent of the distance when launched at 45 degrees. – sebtheiler Feb 27 '18 at 00:14
  • for two angles $\alpha$ and $\beta$, the relationship between distances will be $\frac{\sin (2\alpha)}{\sin (2\beta)}$ – Vasili Feb 27 '18 at 01:08
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Based on your reply to Vasya's answer, is this closer to what you're looking for?

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

Robert Howard
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