Angle and Range of Projectile Motion

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The equation for the range of projectile motion, R = v^2sin(2θ)/g, shows that the angle θ directly influences the distance traveled. Since gravity (g) is constant, the only variable affecting the range is sin(2θ), which reaches its maximum value of 1. This occurs when 2θ equals 90 degrees, leading to θ being 45 degrees. Therefore, a launch angle of 45 degrees yields the maximum distance for a given velocity. This demonstrates the optimal angle for maximizing projectile range.
tharindu
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Using R=v^2sin2x/g how do you prove that the angle of 45 yields the maximum distance?
 
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for a given v (velocity) and since g(the gravity) is constant then the only variable that will affect the range (distance in the x-direction) is sin(2theta) , and the max. number can be obtained from sin(2theta) is 1 (since the range for the sin is between -1 and 1) ..

then you will have sin(2theta) = 1 >> 2theta = arcsin(1) >> 2theta= 90 >> theta = 45 ..

hopefully that was clear .. :)
 
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