A projectile is launched from the ground with an initial velocity of 50 m/s at an angle of 30 degrees above the horizontal. Calculate the maximum height reached by the projectile. Use \( g = 9.8 \, \text{m/s}^2 \).

["# How to Calculate the Maximum Height of a Projectile Launch—A Case Study with 50 m/s at 30°", "When an object is launched into the air, its motion follows a parabolic trajectory governed by physics principles. Understanding key aspects like the launch angle and initial speed allows us to determine important values such as maximum height. In this article, we explore a classic projectile scenario: a projectile launched from ground level with an initial velocity of 50 m/s at a 30° angle above the horizontal. We’ll walk through the calculations to find the maximum height reached, using the standard gravitational acceleration ( g = 9.8 , \ ext{m/s}^2 ).", "## Understanding Projectile Motion Basics", "A projectile launched at an angle follows two independent motions: horizontal (constant velocity) and vertical (affected by gravity). The vertical component of velocity determines the maximum height. At the peak of its trajectory, the vertical velocity becomes zero.", "Given:", "- Initial speed, ( v_0 = 50 , \ ext{m/s} )\n- Launch angle, ( \ heta = 30^\circ )\n- Vertical acceleration, ( g = 9.8 , \ ext{m/s}^2 ) (acting downward)", "## Step 1: Resolve Initial Velocity into Vertical Component", "The vertical component of the initial velocity is:", "[\nv_{0y} = v_0 \sin \ heta = 50 \ imes \sin(30^\circ)\n]", "Since ( \sin(30^\circ) = 0.5 ),", "[\nv_{0y} = 50 \ imes 0.5 = 25 , \ ext{m/s}\n]", "## Step 2: Use Kinematic Equation to Find Maximum Height", "At maximum height, the final vertical velocity ( v_y = 0 ). Using the kinematic equation:", "[\nv_y^2 = v_{0y}^2 - 2 g h_{\ ext{max}}\n]", "Setting ( v_y = 0 ):", "[\n0 = (25)^2 - 2 \ imes 9.8 \ imes h_{\ ext{max}}\n]", "[\n0 = 625 - 19.6 , h_{\ ext{max}}\n]", "Solving for ( h_{\ ext{max}} ):", "[\n19.6 , h_{\ ext{max}} = 625\n]", "[\nh_{\ ext{max}} = \frac{625}{19.6} \approx 31.89 , \ ext{meters}\n]", "## Conclusion", "The projectile launched at 50 m/s and 30° above the horizontal reaches a maximum height of approximately 31.89 meters before descending back to the ground. Understanding this calculation helps in fields ranging from sports science to engineering, where predicting motion trajectories is essential.", "Keywords: projectile motion, maximum height calculation, projectile launched 50 m/s 30 degrees, kinematics, physics calculation, vertical velocity, g = 9.8 m/s², trigonometry in projectile motion.", "---", "This SEO-friendly article explains the concept clearly and guides readers through a step-by-step solution, incorporating relevant keywords to boost search visibility for related physics questions."]