A projectile is launched with an initial velocity of 50 m/s at an angle of 30 degrees. Find the maximum height reached.
["Projectile Launch Angle and Maximum Height: A Physics Problem Explained", "Understanding projectile motion is essential in physics, especially when analyzing objects launched at an angle. In this article, we explore a classic projectile motion problem: a projectile launched with an initial velocity of 50 m/s at a 30-degree angle, focusing on calculating the maximum height reached.", "---", "### The Physics Behind Projectile Launch", "When an object is launched with an initial velocity ( v_0 ) at an angle ( \ heta ) relative to the horizontal, its motion can be broken into horizontal and vertical components. While horizontal motion involves constant velocity (in the absence of air resistance), vertical motion is influenced by gravity, resulting in a symmetrical rise and fall pattern.", "The maximum height ( H ) is the peak vertical position reached by the projectile after its ascent. It depends entirely on the initial vertical component of velocity and the acceleration due to gravity.", "---", "### Step-by-Step: Calculating Maximum Height", "1. Decompose the initial velocity", "Given:\n- Initial speed ( v_0 = 50 , \ ext{m/s} )\n- Launch angle ( \ heta = 30^\circ )", "The vertical component of velocity is:\n[\nv_{0y} = v_0 \sin \ heta = 50 \ imes \sin 30^\circ = 50 \ imes 0.5 = 25 , \ ext{m/s}\n]", "2. Use the kinematic equation for vertical motion", "At the maximum height, the vertical velocity becomes zero (( v_y = 0 )) due to gravitational acceleration (( g = 9.8 , \ ext{m/s}^2 ) downward).", "Using the formula:\n[\nv_y^2 = v_{0y}^2 - 2 g H\n]\nSet ( v_y = 0 ):\n[\n0 = (25)^2 - 2 \ imes 9.8 \ imes H\n]\n[\n625 = 19.6 \ imes H\n]\n[\nH = \frac{625}{19.6} \approx 31.89 , \ ext{meters}\n]", "---", "### Why This Matters", "Knowing the maximum height helps in various applications—from launching sports equipment to ballistic modeling. It demonstrates how the launch angle influences vertical reach, even when the horizontal speed is high. With ( \ heta = 30^\circ ), this launch achieves a balance between horizontal distance and vertical climb, maximizing altitude at 50 m/s.", "---", "### Final Answer", "The maximum height reached by the projectile launched at 50 m/s at 30 degrees is approximately 31.9 meters.", "---", "### Key Formula Recap", "[\nH = \frac{(v_0 \sin \ heta)^2}{2g}\n]\nPlugging in values confirms:\n[\nH = \frac{(50 \sin 30^\circ)^2}{2 \ imes 9.8} = \frac{(25)^2}{19.6} \approx 31.89 , \ ext{m}\n]", "---", "For anyone studying physics or engineering, mastering projectile calculations like this provides the foundation for understanding motion, trajectories, and real-world applications—from soccer kicks to spacecraft launches."]
