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From the above equation, show that for y = y0, maximum range is achieved at θ = 45◦

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Final answer:

The maximum range in projectile motion, when air resistance is negligible, is achieved at a launch angle of 45° because the sine function in the range formula reaches its maximum at this angle.

Step-by-step explanation:

Understanding Projectile Motion for Maximum Range

To show that the maximum range in projectile motion is achieved at a launch angle of θ = 45°, we can look at the physics behind the projectile motion without air resistance. For a given initial velocity, the range (R) of a projectile is given by the formula R = (v^2 × sin(2θ)) / g, where v is the initial velocity, g is the acceleration due to gravity, and θ is the launch angle. The function sin(2θ) reaches its maximum value of 1 when 2θ equals 90°, which means that θ equals 45° for the maximum range. Furthermore, for two launch angles that add up to 90°, the projectile will have the same range. The lower angle will result in a shorter apex, while the higher angle will result in a higher apex.

The concept that the maximum range is achieved at 45° is valid only when air resistance is negligible. When air resistance is considered, the optimal launch angle is slightly less than 45°. However, for simple projectile motion calculations where air resistance is not a factor, setting the angle to 45° provides the longest range possible for a given initial velocity.

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