In ideal projectile motion on level ground (ignore air resistance), a firearm fires a projectile with fixed speed $v$ from ground level. Which statement about the projectile’s maximum horizontal range is true?

• A. The maximum range occurs when the launch angle is $30°$.
• B. The range is the same for all launch angles because gravity acts only vertically.
• C. The maximum range occurs for a horizontal launch, $0°$.
• D. The maximum range occurs when the launch angle is $45°$.

Answer: (D)

Verified step by step guidance

1. Recall the formula for the horizontal range \(R\) of a projectile launched from ground level with initial speed \(v\) at an angle \(\theta\) to the horizontal: \[R = \frac{v^{2} \sin(2\theta)}{g}\] where \(g\) is the acceleration due to gravity.
2. Understand that the range depends on the term \(\sin(2\theta)\), which varies with the launch angle \(\theta\).
3. To find the angle that maximizes the range, consider the function \(\sin(2\theta)\) and determine its maximum value. Since \(\sin\) function reaches its maximum value of 1 at \$90^\circ$, set: \[2\theta = 90^\circ \implies \theta = 45^\circ\]
4. Interpret this result: the maximum horizontal range occurs when the launch angle is \$45^\circ$ because this angle maximizes the product of the horizontal and vertical components of the initial velocity, optimizing the distance traveled.
5. Conclude that the other options are incorrect because: - At \$30^\circ\(, \)\sin(60^\circ)\( is less than 1, so range is not maximum. - The range is not the same for all angles since \)\sin(2\theta)\( changes with \)\theta\(. - A horizontal launch (\)0^\circ$) results in zero vertical motion, so the projectile immediately hits the ground, giving zero range.