A projectile is launched with an initial velocity of 50 m/s at an angle of 30°. Find the maximum height reached.
["# Finding the Maximum Height of a Projectile Launched at 50 m/s at 30°", "When a projectile is launched with an initial velocity and at a specific launch angle, understanding its maximum height is a key topic in physics and engineering. In this article, we’ll explore how to calculate the maximum height reached by a projectile launched at 50 meters per second (m/s) at an angle of 30° above the horizontal. This calculation involves principles of kinematics and vector decomposition, making it essential for students, educators, and physics enthusiasts.", "## Understanding Projectile Motion", "Projectile motion occurs when an object is launched into the air and moves under the influence of gravity alone—ignoring air resistance. The initial velocity can be broken into two components: horizontal and vertical. The vertical component determines how high the projectile rises before descending back to the ground.", "Once the projectile reaches its peak, its vertical velocity momentarily becomes zero before acceleration due to gravity begins pulling it back downward. Therefore, the maximum height is solely a function of the vertical motion.", "## Step-by-Step Calculation", "### 1. Decompose Initial Velocity into Components", "Given:\n- Initial speed ( v_0 = 50 , \ ext{m/s} )\n- Launch angle ( \ heta = 30^\circ )", "The vertical component of the initial velocity is:\n[\nv_{y0} = v_0 \sin \ heta = 50 \ imes \sin 30^\circ = 50 \ imes 0.5 = 25 , \ ext{m/s}\n]", "### 2. Use Kinematic Equation for Maximum Height", "At maximum height, the vertical velocity becomes 0. Using the kinematic equation:\n[\nv_y^2 = v_{y0}^2 - 2g h_{\ ext{max}}\n]\nwhere:\n- ( v_y = 0 ) (at the peak)\n- ( g = 9.8 , \ ext{m/s}^2 ) (acceleration due to gravity)\n- ( h_{\ ext{max}} ) is the maximum height", "Rearranging to solve for ( h_{\ ext{max}} ):\n[\n0 = v_{y0}^2 - 2g h_{\ ext{max}} \implies h_{\ ext{max}} = \frac{v_{y0}^2}{2g}\n]", "### 3. Plug in the Values", "[\nh_{\ ext{max}} = \frac{(25)^2}{2 \ imes 9.8} = \frac{625}{19.6} \approx 31.89 , \ ext{meters}\n]", "## Summary", "By calculating the vertical component of the initial velocity and applying the fundamental kinematic formula, we find that a projectile launched at 50 m/s and 30° reaches a maximum height of approximately 31.89 meters. This result highlights the importance of vector decomposition and gravitational acceleration in predicting projectile motion.", "## Frequently Asked Questions", "Q: Why do we only consider the vertical component for maximum height?\nA: Because although horizontal motion continues uniformly (neglecting air resistance), vertical motion is entirely governed by gravity, which brings the projectile to a halt at its peak.", "Q: How does a higher launch angle affect the maximum height?\nA: A higher launch angle increases the vertical component of velocity, resulting in a greater maximum height—up to 90°, where the projectile is launched straight up and hits its maximum height exactly at launch.", "Q: Does air resistance play a role?\nA: In ideal projectile motion problems, air resistance is ignored. In real-world scenarios, air resistance reduces height and shortens range, but this calculation assumes a vacuum.", "By mastering these principles, you gain a deeper understanding of projectile dynamics—critical for fields like sports science, engineering, and astronomy."]
