A projectile is launched with velocity \(v_0 = 30 \, \text{m/s}\) at an angle of \(45^\circ\). Find the maximum height reached.
["# Projectile Motion: Maximum Height When Launched at 45° with Initial Velocity 30 m/s", "When a projectile is launched into the air, its motion is governed by both horizontal and vertical components, with gravity acting downward. Understanding key aspects like maximum height is essential in physics and engineering. This SEO-optimized article explains how to find the maximum height reached by a projectile launched at (v_0 = 30 , \ ext{m/s}) at an angle of (45^\circ).", "## Understanding Projectile Motion", "Projectile motion occurs when an object is thrown or launched under the influence of gravity alone, ignoring air resistance. The trajectory is a parabola, and the vertical motion determines the height reached. The vertical component of the initial velocity governs how high the projectile ascends before falling back.", "## Step 1: Break the Initial Velocity into Components", "The initial velocity (v_0 = 30 , \ ext{m/s}) is launched at (45^\circ) above the horizontal.", "- Horizontal component:\n [\n v_{0x} = v_0 \cos(45^\circ) = 30 \ imes \frac{\sqrt{2}}{2} \approx 21.21 , \ ext{m/s}\n ]", "- Vertical component:\n [\n v_{0y} = v_0 \sin(45^\circ) = 30 \ imes \frac{\sqrt{2}}{2} \approx 21.21 , \ ext{m/s}\n ]", "## Step 2: Determine Time to Reach Maximum Height", "At the peak of its trajectory, the vertical velocity becomes zero. The time to reach maximum height is found using:", "[\nv_y = v_{0y} - gt\n]", "At maximum height, (v_y = 0):", "[\n0 = 21.21 - 9.8t \quad \Rightarrow \quad t = \frac{21.21}{9.8} \approx 2.16 , \ ext{seconds}\n]", "## Step 3: Calculate Maximum Height Using Kinematic Equation", "Use the vertical displacement formula:", "[\ny = v_{0y} t - \frac{1}{2} g t^2\n]", "Substitute values:", "[\ny = (21.21)(2.16) - \frac{1}{2}(9.8)(2.16)^2\n]", "Calculate:", "[\ny \approx 45.78 - 22.88 \approx 22.9 , \ ext{meters}\n]", "## Summary", "When a projectile is launched at (30 , \ ext{m/s}) at (45^\circ), the maximum height it reaches is approximately 22.9 meters. This result combines trigonometric velocity components and basic kinematics for precise vertical motion analysis.", "## Keywords for SEO Optimization:\nprojectile motion maximum height formula, projectile launched at 45 degrees, vertical component of velocity, maximum height calculation, kinematics projectile motion, physics projectile height formula, how to find max height in projectile motion", "---", "This article balances clear explanations with strategic keyword inclusion to maximize visibility on search engines while delivering valuable physics insights for students, educators, and enthusiasts."]
