A projectile is launched with an initial velocity of 50 m/s at an angle of 30 degrees. What is the horizontal range of the projectile? (Assume \( g = 9.8 \, \text{m/s}^2 \) and ignore air resistance.)

["# Projectile Motion: Calculating the Horizontal Range When Launched at 50 m/s at 30°", "Understanding projectile motion is fundamental in physics, especially in fields like engineering, sports science, and ballistics. A common question students encounter is: What is the horizontal range of a projectile launched at 50 m/s at a 30-degree angle? In this article, we’ll break down the physics behind this calculation, explain how to determine the range, and provide a clear, step-by-step solution with practical insights.", "## What Is a Projectile?", "A projectile is any object launched into the air that is subject only to the acceleration due to gravity—air resistance is neglected here. Once launched, the motion can be analyzed using two independent components: horizontal motion and vertical motion. Horizontal velocity remains constant (ignoring air resistance), while vertical motion is affected downward by gravity.", "## Key Concepts for Horizontal Range", "- Initial velocity ((v_0)): 50 m/s\n- Launch angle ((\ heta)): 30°\n- Acceleration due to gravity ((g)): (9.8 , \ ext{m/s}^2) (downward)\n- Range ((R)): Horizontal distance traveled before hitting the ground", "The horizontal range depends primarily on:\n- The horizontal component of initial velocity ((v_{0x}))\n- The time ((t)) the projectile remains in the air", "Because horizontal velocity is constant and vertical motion determines flight time, the formula for horizontal range is:", "[\nR = v_{0x} \cdot t\n]", "where\n( v_{0x} = v_0 \cdot \cos(\ heta) )\n( t = \frac{2v_0 \sin(\ heta)}{g} )", "### Why use flight time?\nSince the projectile goes up and then comes down, we calculate total time in the air as twice the time to reach maximum height. At peak height, vertical velocity becomes zero.", "## Step-by-Step Calculation", "### Step 1: Compute the horizontal component of velocity", "[\nv_{0x} = v_0 \cdot \cos(\ heta) = 50 \cdot \cos(30^\circ)\n]", "We know that (\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866)", "[\nv_{0x} = 50 \cdot 0.866 = 43.3 , \ ext{m/s}\n]", "### Step 2: Compute the total time of flight", "The full time of flight is given by:", "[\nt = \frac{2v_0 \sin(\ heta)}{g}\n]", "Compute (\sin(30^\circ)):\n(\sin(30^\circ) = 0.5)", "[\nt = \frac{2 \cdot 50 \cdot 0.5}{9.8} = \frac{50}{9.8} \approx 5.10 , \ ext{seconds}\n]", "### Step 3: Compute the horizontal range", "[\nR = v_{0x} \cdot t = 43.3 \cdot 5.10\n]", "Using multiplication:", "[\nR \approx 220.83 , \ ext{meters}\n]", "## Final Result", "The horizontal range of the projectile launched at 50 m/s at a 30° angle is approximately 220.8 meters.", "## Why This Formula Works", "The horizontal motion is uniform (constant velocity), while vertical motion is accelerated. Breaking the motion into components allows us to independently compute time in the air and distance covered horizontally. Ignoring air resistance simplifies the calculation and provides an approximate but reliable result for most real-world applications involving short-range trajectories.", "---", "Summary\n- Initial velocity: 50 m/s\n- Angle: 30°\n- Horizontal range ≈ 220.8 meters\n- Formula: ( R = v_0 \cos(\ heta) \cdot \frac{2v_0 \sin(\ heta)}{g} = \frac{v_0^2 \sin(2\ heta)}{g} ) (alternative direct formula)", "For those applying this to sports or engineering, understanding projectile range helps optimize targets, launches, and performance predictions.", "---", "Keywords: projectile motion, horizontal range, physics projectile, initial velocity 50 m/s, angle 30 degrees, time of flight, gravity, projectile trajectory, range formula, constant velocity."]