A projectile is launched with an initial velocity of 20 m/s at an angle of 30 degrees. Find the maximum height reached.
["# Projectile Motion: How to Calculate Maximum Height When Launched at 20 m/s at 30 Degrees", "Understanding projectile motion is essential in physics, especially in fields like sports, engineering, and aerospace. One common question students ask is: What is the maximum height a projectile reaches when launched with an initial velocity of 20 m/s at a 30-degree angle? This article explains how to calculate the maximum height using fundamental physics principles.", "## What Is Projectile Motion?", "Projectile motion describes the motion of an object launched into the air, subject only to gravity. Once the projectile is in flight, its horizontal and vertical motions are independent of each other. The vertical component determines the height achieved, making it the key factor when calculating the maximum height.", "---", "## Given Values", "- Initial velocity (( v_0 )) = 20 m/s\n- Launch angle (( \ heta )) = 30°\n- Acceleration due to gravity (( g )) = 9.8 m/s² (downward)", "---", "## Step-by-Step Calculation", "### 1. Break the initial velocity into vertical and horizontal components", "The vertical component of velocity determines how high the projectile goes. It is calculated using:", "[\nv_y = v_0 \sin(\ heta)\n]", "Substituting the given values:", "[\nv_y = 20 \ imes \sin(30^\circ)\n]", "Since ( \sin(30^\circ) = 0.5 ),", "[\nv_y = 20 \ imes 0.5 = 10 \ ext{ m/s (upward)}\n]", "---", "### 2. Use the kinematic equation to find maximum height", "At the maximum height, the vertical velocity becomes 0 (instantarily stops going up, then starts falling down). We use the equation:", "[\nv_y^2 = v_{y,0}^2 - 2gH\n]", "Where:\n- ( v_y ) = final vertical velocity (0 m/s at the top)\n- ( v_{y,0} ) = initial vertical velocity = 10 m/s\n- ( H ) = maximum height we want to find\n- ( g ) = 9.8 m/s²", "Rearranging the equation:", "[\n0 = (10)^2 - 2 \ imes 9.8 \ imes H\n]", "[\n0 = 100 - 19.6H\n]", "[\n19.6H = 100\n]", "[\nH = \frac{100}{19.6} \approx 5.10 \ ext{ meters}\n]", "---", "## Conclusion", "The maximum height reached by a projectile launched at 20 m/s at a 30-degree angle is approximately 5.10 meters. This result is derived from breaking the initial velocity into vertical components and applying basic kinematic equations under constant gravitational acceleration. Understanding this process helps solve real-world problems involving projectiles, such as launching sports equipment or ballistic calculations.", "---", "## Key Takeaways", "- The vertical component of initial velocity governs maximum height.\n- Use ( v_y = v_{y,0} - gt ) or the kinematic equation ( v_y^2 = v_{y,0}^2 - 2gH ) to find peak altitude.\n- Angle of launch directly affects vertical velocity and thus height.\n- Precision matters: using ( g = 9.8 , \ ext{m/s}^2 ) ensures accurate results.", "---", "**Keywords: projectile motion, maximum height, physics formula, velocity components, kinematic equations, projects with velocity 20 m/s, 30 degree launch angle, calculate height, gravity effects", "---", "Whether you're studying physics, coaching sports, or building models, mastering projectile height calculations empowers accurate predictions and deeper understanding of motion."]
