Neglecting air resistance, a projectile launched at which angle to the horizontal has the greatest range on level ground?

Introduction / Context:
Projectile motion is a standard topic in kinematics. For a given initial speed and launch from and to the same height, the horizontal range depends on the angle of projection. This question asks you to identify the angle that gives the maximum range on level ground when air resistance is neglected. Knowing this result helps with many practical problems, such as maximising distance in sports or ballistics under ideal conditions.

Given Data / Assumptions:
- The projectile is launched with fixed initial speed from ground level and lands back at the same level.
- Air resistance is neglected, so only uniform gravitational acceleration acts vertically.
- The options are 30 degrees, 45 degrees, 60 degrees and 90 degrees.

Concept / Approach:
The horizontal range R of a projectile launched with initial speed u at an angle theta to the horizontal (from and to the same level) is given by the formula R = (u^2 * sin(2 * theta)) / g, where g is acceleration due to gravity. For fixed u and g, the range depends only on sin(2 * theta). The maximum value of sin(2 * theta) is 1, which occurs when 2 * theta = 90 degrees. This means theta = 45 degrees gives the maximum range under these ideal conditions.

Step-by-Step Solution:
Step 1: Use the range formula: R = (u^2 * sin(2 * theta)) / g for equal launch and landing heights. Step 2: For fixed u and g, maximise the term sin(2 * theta) to maximise R. Step 3: The maximum value of sine of any angle is 1, so sin(2 * theta) is maximum when 2 * theta = 90 degrees. Step 4: Solve for theta: 2 * theta = 90 degrees implies theta = 45 degrees. Step 5: Therefore, a launch angle of 45 degrees gives the greatest horizontal range. Step 6: Check the options and select 45 degrees as the correct answer.

Verification / Alternative check:
You can compare the ranges at other angles using the same formula. For 30 degrees, 2 * theta = 60 degrees and sin(60 degrees) is about 0.866, so the range is less than the maximum. For 60 degrees, 2 * theta = 120 degrees and sin(120 degrees) is also about 0.866, giving the same range as 30 degrees, still less than the maximum. At 90 degrees, 2 * theta = 180 degrees and sin(180 degrees) is zero, so the range is zero, meaning the projectile goes straight up and comes straight down. This confirms that 45 degrees is indeed the peak point for range.

Why Other Options Are Wrong:
30 degrees: Gives a good range but not the maximum since sin(60 degrees) is less than 1.
60 degrees: Produces the same range as 30 degrees because sin(120 degrees) equals sin(60 degrees), again less than the maximum possible value of 1.
90 degrees: The projectile rises and falls vertically, giving zero horizontal range, which is clearly not maximum.

Common Pitfalls:
Some learners think that a smaller angle like 30 degrees or a larger angle like 60 degrees must give more distance because they imagine flatter trajectories or higher arcs. The key is to remember that the mathematical combination of vertical and horizontal components is optimal at 45 degrees. Another mistake is to ignore the assumption of no air resistance. In real life, with drag and other effects, the best angle can be slightly less than 45 degrees, but in ideal textbook problems, 45 degrees is the correct answer.

Final Answer:
Neglecting air resistance, a projectile has the greatest range when launched at 45 degrees to the horizontal.