\(h = ut + \frac{1}{2}at^2 = 30 \times 3.06 - \frac{1}{2} \times 9.8 \times (3.06)^2 \approx 45.9 \, \text{meters}\)

["Understanding Projectile Motion: Calculating Maximum Height Using ( h = ut + \frac{1}{2}at^2 )", "When analyzing projectile motion, one of the most fundamental calculations involves determining the maximum height an object reaches under gravity. This becomes particularly interesting when solving equations like:", "[\nh = ut + \frac{1}{2}at^2\n]", "In this example, we apply this formula with real-world values to illustrate how physics and algebra combine to explain motion.", "---", "### What is the Formula ( h = ut + \frac{1}{2}at^2 )?", "In projectile motion, the term ( h ) represents the height of an object at time ( t ). The equation combines two components:", "- ( ut ): The initial vertical velocity multiplied by time (this accounts for upward motion under acceleration).\n- ( \frac{1}{2}at^2 ): The displacement due to constant acceleration (here due to gravity acting downward).", "Since gravity opposes upward motion, acceleration ( a = -9.8 , \ ext{m/s}^2 ), so the full expression becomes:", "[\nh = ut + \frac{1}{2}(-9.8)t^2 = ut - 4.9t^2\n]", "With correct sign conventions and correct coefficients, it simplifies conveniently to:", "[\nh = ut + \frac{1}{2}at^2\n]", "---", "### Applying the Values", "Let’s plug in real values from the classic example:\nInitial velocity ( u = 30 , \ ext{m/s} ),\nTime ( t = 3.06 , \ ext{seconds} ),\nGravitational acceleration ( a = -9.8 , \ ext{m/s}^2 ).", "Compute each term:", "1. First term: ( ut = 30 \ imes 3.06 = 91.8 , \ ext{meters} )\nThis represents the height gained purely from initial upward speed.", "2. Second term: ( \frac{1}{2}at^2 = 0.5 \ imes (-9.8) \ imes (3.06)^2 )\nCalculate step-by-step:\n[\n(3.06)^2 = 9.3636\n]\n[\n0.5 \ imes (-9.8) = -4.9\n]\n[\n-4.9 \ imes 9.3636 \approx -45.9 , \ ext{meters}\n]", "3. Combine both terms:\n[\nh \approx 91.8 - 45.9 = 45.9 , \ ext{meters}\n]", "---", "### Final Result", "Rounding appropriately, the maximum height reached is approximately 45.9 meters.", "---", "### Why This Matters", "Understanding this calculation helps in disciplines like ballistics, sports physics, and engineering. The balance between initial upward velocity and gravitational acceleration determines the peak height — a concept essential for anything from modeling cannon fire to predicting the arc of a basketball shot.", "---", "### Key Takeaways", "- Use ( h = ut + \frac{1}{2}at^2 ) to model vertical motion under constant acceleration.\n- Acceleration due to gravity acts downward and is negative in upward-moving frames.\n- Proper sign handling and arithmetic are crucial to accurate results.\n- This formula enables prediction of trajectories in practical applications.", "---", "### Related SEO Keywords", "- projectile motion formula\n- height of a projectile calculation\n- ( h = ut + \frac{1}{2}at^2 ) physics\n- maximum height under gravity\n- projectile motion explanation\n- physics formula application examples", "---", "### Summarized Equation\n[\n\boxed{h \approx 45.9 , \ ext{meters}}\n]\nValid for initial velocity ( u = 30 , \ ext{m/s} ) at ( t = 3.06 , \ ext{s} ), ( a = -9.8 , \ ext{m/s}^2 )", "---", "Dive deeper into the fascinating world of motion — physics formulas are not just abstract concepts, but key tools to unlock real-world dynamics!"]