A projectile is launched with an initial velocity of 50 m/s at an angle of 30 degrees to the horizontal. What is the horizontal range of the projectile? (Assume no air resistance and \( g = 9.8 \, \text{m/s}^2 \))

["Projectile Motion: Calculating the Horizontal Range of a 50 m/s Launch at 30 Degrees", "Understanding projectile motion is fundamental in physics and engineering, helping to predict the trajectory of objects like sports balls, missiles, and aerospace vehicles. In this article, we explore a classic projectile problem: determining the horizontal range of a projectile launched at 50 meters per second at a 30-degree angle with no air resistance and gravitational acceleration of ( g = 9.8 , \ ext{m/s}^2 ).", "---", "### Key Parameters of the Projectile", "- Initial velocity (( v_0 )): 50 m/s\n- Launch angle (( \ heta )): 30 degrees\n- Acceleration due to gravity (( g )): 9.8 m/s² (downward)\n- Air resistance: Neglected", "---", "### Understanding Horizontal Range in Projectile Motion", "The horizontal range ( R ) is the total horizontal distance the projectile travels before hitting the ground. It depends on two main components:", "- The horizontal component of velocity (( v_{0x} )), which remains constant (no air resistance).\n- The vertical motion, which determines the time the projectile stays in the air.", "---", "### Step 1: Resolve Initial Velocity into Components", "Horizontal:\n[\nv_{0x} = v_0 \cdot \cos(\ heta) = 50 \cdot \cos(30^\circ)\n]\nSince ( \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 ),\n[\nv_{0x} = 50 \cdot 0.866 = 43.3 , \ ext{m/s}\n]", "Vertical:\n[\nv_{0y} = v_0 \cdot \sin(\ heta) = 50 \cdot \sin(30^\circ)\n]\nSince ( \sin(30^\circ) = 0.5 ),\n[\nv_{0y} = 50 \cdot 0.5 = 25 , \ ext{m/s}\n]", "---", "### Step 2: Determine Time of Flight", "The total time in the air depends on the vertical motion. The projectile rises to its peak, pauses, and then falls back to the ground. The time to reach the maximum height is:\n[\nt_{\ ext{up}} = \frac{v_{0y}}{g} = \frac{25}{9.8} \approx 2.55 , \ ext{seconds}\n]", "The total time of flight (( t_{\ ext{total}} )) is twice this since the time up equals time down:\n[\nt_{\ ext{total}} = 2 \cdot t_{\ ext{up}} = 2 \cdot 2.55 = 5.1 , \ ext{seconds}\n]", "---", "### Step 3: Calculate Horizontal Range", "Since horizontal velocity is constant,\n[\nR = v_{0x} \cdot t_{\ ext{total}}\n]\nSubstitute values:\n[\nR = 43.3 , \ ext{m/s} \ imes 5.1 , \ ext{s} \approx 220.83 , \ ext{meters}\n]", "---", "### Final Answer:\nThe horizontal range of the projectile is approximately 220.8 meters.", "---", "### Why Knowing Projectile Range Matters", "From sports to military applications and aerospace engineering, calculating the range helps design trajectories, optimize performance, and improve accuracy. This simple yet powerful analysis shows how initial velocity and launch angle determine how far and where a projectile lands.", "---", "### Summary", "- Initial velocity: 50 m/s at 30°\n- Horizontal component: 43.3 m/s\n- Time of flight: ~5.1 seconds\n- Horizontal range: ~220.8 meters", "Always remember: without air resistance, and with a symmetric launch, the horizontal range formula simplifies elegantly to ( R = \frac{v_0^2 \sin(2\ heta)}{g} ), but breaking it into components gives clear insight into motion physics.", "---", "Key SEO keywords: projectile motion, horizontal range formula, projectile launch angle 30 degrees, projectile velocity 50 m/s, range calculation, constant velocity motion, physics projectile range, gravity effect on trajectory, no air resistance projectile, physics problem solving."]